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Question

Bài 3 : Tìm x, biết :

a) 16x^2 - (4x - 5)^2 = 15

b) (2x + 3)^2 - 4×(x - 1)×(x+ 1)=49

c) (2x + 1)×(1 - 2x)+(1 - 2x)^2=18

d) 2×(x + 1)^2 - (x - 3)×(x + 3) - (x - 4)^2=0

e) (x -5)^2 - x×(x - 4)=9

f...

Lê Trang · Accepted Answer

Bài 3: Tìm x, biết:

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

$\Leftrightarrow16x^2-16x^2+40x-25-15=0$

$\Leftrightarrow40x-40=0$

$\Leftrightarrow4x=40$

$\Leftrightarrow x=10$

Vậy x = 10

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

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

$\Leftrightarrow4x^2+12x+9-4x^2+4-49=0$

$\Leftrightarrow12x-36=0$

$\Leftrightarrow12x=36$

$\Leftrightarrow x=3$

Vậy x = 3

c) $\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18$

$\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18$

$\Leftrightarrow2\left(1-2x\right)=18$

$\Leftrightarrow2-4x=18$

$\Leftrightarrow4x=-16$

$\Leftrightarrow x=-4$

Vậy x =-4

d) $2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0$

$\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0$

$\Leftrightarrow12x-5=0$

$\Leftrightarrow12x=5$

$\Leftrightarrow x=\frac{5}{12}$

Vậy $x=\frac{5}{12}$

e) $\left(x-5\right)^2-x\left(x-4\right)=9$

$\Leftrightarrow x^2-10x+25-x^2+4x=9$

$\Leftrightarrow25-6x=9$

$\Leftrightarrow6x=16$

$\Leftrightarrow x=\frac{8}{3}$

Vậy $x=\frac{8}{3}$

f) $\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0$

$\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0$

$\Leftrightarrow21-5x=0$

$\Leftrightarrow5x=21$

$\Leftrightarrow x=\frac{21}{5}$

Vậy $x=\frac{21}{5}$Câu trả lời: Bài 3: Tìm x, biết: