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2次式の因数分解/2次方程式の解

Authored by E Anomuder

Mathematics

6th - 8th Grade

Used 3+ times

2次式の因数分解/2次方程式の解
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12 questions

Show all answers

1.

MATH RESPONSE QUESTION

2 mins • 2 pts

因数分解せよ。

12a2b−6ab2−4ab12a^2b-6ab^2-4ab

2ab(6a−3b−2)2ab\left(6a-3b-2\right)

Mathematical Equivalence

ON

2.

MATH RESPONSE QUESTION

2 mins • 2 pts

展開せよ。

(4x−5)2\left(4x-5\right)^2

16x2−40x+2516x^2-40x+25

Mathematical Equivalence

OFF

3.

MATH RESPONSE QUESTION

2 mins • 2 pts

因数分解せよ。

x2+10x−39x^2+10x-39

(x+13)(x−3)\left(x+13\right)\left(x-3\right)

Mathematical Equivalence

ON

4.

MATH RESPONSE QUESTION

2 mins • 2 pts

因数分解せよ。

y2−18y+81y^2-18y+81

(y−9)2\left(y-9\right)^2

Mathematical Equivalence

OFF

5.

MATH RESPONSE QUESTION

2 mins • 2 pts

因数分解せよ。

abx2−16ababx^2-16ab

ab(x+4)(x−4)ab\left(x+4\right)\left(x-4\right)

Mathematical Equivalence

OFF

6.

MATH RESPONSE QUESTION

2 mins • 2 pts

因数分解せよ。

49x2−28xy+4y249x^2-28xy+4y^2

(7x−2y)2\left(7x-2y\right)^2

Mathematical Equivalence

OFF

7.

MATH RESPONSE QUESTION

2 mins • 1 pt

因数分解せよ。

(x+5)2+10(x+5)+16\left(x+5\right)^2+10\left(x+5\right)+16

(x+7)(x+13)\left(x+7\right)\left(x+13\right)

Mathematical Equivalence

ON

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