Ułamek 1 1 − 2 \frac{1}{1-\sqrt{2}} 1 − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 jest równy (usuń niewymierność):

− 1 − 2 -1-\sqrt{2} − 1 − 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 + 2 1+\sqrt{2} 1 + 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 + 2 3 \frac{1+\sqrt{2}}{3} 3 1 + 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 + 3 3 − 2 2 \frac{1+\sqrt{3}}{3-2\sqrt{2}} 3 − 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> 1 + 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Zbiorem rozwiązań nierówności 5 x ( x − 1 ) − 8 x < ( x − 3 ) ( 5 x + 2 ) 5x\left(x-1\right)-8x<\left(x-3\right)\left(5x+2\right) 5 x ( x − 1 ) − 8 x < ( x − 3 ) ( 5 x + 2 ) jest przedział:

x ∈ ( − 6 ; ∞ ) x\in\left(-6;\infty\right) x ∈ ( − 6 ; ∞ )

x ∈ ( − ∞ ; − 6 ) x\in\left(-\infty;-6\right) x ∈ ( − ∞ ; − 6 )

Język matematyki

Authored by Magdalena Sibierska

Mathematics

1st - 5th Grade

Used 307+ times

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Język matematyki

Suma zbiorów A = {1,2,3,4,5} i B= {1,3,7,8}, to zbiór X=

Jeśli A=⟨−8,12⟩ i B=(0,20), to różnica A∖B jest przedziałem:

( -2; 3) ∩ 〈 1; 5〉

Wyrażenie ( x + 2) ( y - 3 ) po wymnożeniu ma postać :

Mnożąc jednomian przez sumę algebraiczną 3x(x-7) otrzymasz:

Wyrażenie (2x−y)2\left(2x-y\right)^2(2x−y)2 jest równe:

Wyłączając przed nawias czynnik liczbowy oraz wszystkie możliwe zmienne w jak największych potęgach z 21a^4b^4-27a^2b^2+6a^2b^521a4b4−27a2b2+6a2b5

Ułamek 11−2\frac{1}{1-\sqrt{2}}1−2​1​ jest równy (usuń niewymierność):

Wynik mnożenia √(√2+1)⋅√(√2−1)=√(√2+1)⋅√(√2-1)=√(√2+1)⋅√(√2−1)= wynosi (a)

Które liczby spełniają podaną nierówność 9(x+1)2−(3x−1)2≥−49\left(x+1\right)^2-\left(3x-1\right)^2\ge-49(x+1)2−(3x−1)2≥−4

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Mnożąc jednomian przez sumę algebraiczną
3x(x-7) otrzymasz:

Wyrażenie $\left(2x-y\right)^2$ jest równe:

Wyłączając przed nawias czynnik liczbowy oraz wszystkie możliwe zmienne w jak największych potęgach z $21a^4b^4-27a^2b^2+6a^2b^5$

Ułamek $\frac{1}{1-\sqrt{2}}$ jest równy (usuń niewymierność):

Wynik mnożenia $√(√2+1)⋅√(√2-1)=$ wynosi

(a)

Które liczby spełniają podaną nierówność
$9\left(x+1\right)^2-\left(3x-1\right)^2\ge-4$