Câu hỏi của Lê Thị Xuân Niên

Question

Tìm x :

a ) $16x^2-\left(4x-5\right)^2=15$

b ) $\left(2x+3\right)^2$$-4.\left(x-1\right).\left(x+1\right)=49$

Diễm Quỳnh · Accepted Answer

a)$16x^2-\left(4x-5\right)^2=15$

$\Rightarrow\left(4x\right)^2-\left(4x-5\right)^2-15=0$

$\Rightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)-15=0$

$\Rightarrow5\left(8x-5\right)-15=0$

$\Rightarrow40x-25-15=0$

$\Rightarrow40x-40=0$

$\Rightarrow x=1$Câu trả lời: a)$16x^2-\left(4x-5\right)^2=15$

$\Rightarrow\left(4x\right)^2-\left(4x-5\right)^2-15=0$

$\Rightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)-15=0$

$\Rightarrow5\left(8x-5\right)-15=0$

$\Rightarrow40x-25-15=0$

$\Rightarrow40x-40=0$

$\Rightarrow x=1$

Diễm Quỳnh · Answer

b)$\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49$

$\Rightarrow\left(4x^2-12x-9\right)-4\left(x^2-1\right)-49=0$

$\Rightarrow4x^2-12x+9-4x^2+4-49=0$

$\Rightarrow12x-36=0$

$\Rightarrow12\left(x-3\right)=0$

$\Rightarrow x=3$Câu trả lời: b)$\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49$

$\Rightarrow\left(4x^2-12x-9\right)-4\left(x^2-1\right)-49=0$

$\Rightarrow4x^2-12x+9-4x^2+4-49=0$

$\Rightarrow12x-36=0$

$\Rightarrow12\left(x-3\right)=0$

$\Rightarrow x=3$

Bài 3: Những hằng đẳng thức đáng nhớ