a,(2x-1)^2=49

<=>(2x-1)^2=7^2

<=>(2x-1)^2-7^2=0

<=>(2x-1-7)(2x-1+7)=0

<=>(2x-8)(2x+6)=0

<=>2x-8=0 hoặc 2x+6=0

<=>x=4 hoặc x=-3

a) (2x - 1)² - 49 = 0

⇔ (2x - 1 + 7)(2x - 1 - 7) = 0

⇔ (2x + 6)(2x - 8) = 0

⇔\(\left[{}\begin{matrix}2x+6=0\\2x-8=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

Vậy nghiệm của pt là x = -3 và x = 4

b) (x + 2)² - [3(x - 2)]² = 0

⇔ (x + 2 - 3x + 6)(x + 2 + 3x - 6) = 0

⇔(-2x + 8)(4x - 4) = 0

⇔ \(\left[{}\begin{matrix}-2x+8=0\\4x-4=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

Vậy nghiệm của pt là x = 4; x = 1

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a) \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

jup mk mấy câu kia vs

b, \(\left(x+2\right)^2=9\left(x-4x+4\right)\)

\(\Leftrightarrow\left(x+2\right)^2=\left[3\left(x-2\right)\right]^2\)

\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)

\(\Leftrightarrow\left(4x-4\right)\left(-2x+8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

\(a,\left(2x-1\right)^2=49\)

\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(4x^2+28x+49=9x^2+36x+36\)

\(4x^2+28x+49-9x^2-36x-36=0\)

\(-5x^2-8x+13=0\)

\(5x^2+13-5x-13=0\)

\(x\left(5x+13\right)-1\left(5x+13\right)=0\)

\(\left(x-1\right)\left(5x+13\right)=0\)

\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)

\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(x=-5\)

\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)

\(25x^2-30x+9-16x^2+56x-49=0\)

\(9x^2+26x-40=0\)

\(9x^2+36x-10x-40=0\)

\(9x\left(x+4\right)-10\left(x+4\right)=0\)

\(\left(9x-10\right)\left(x+4\right)=0\)

\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)

e)

\(\left(x+2\right)^2=9\cdot\left(x^2-4x+4\right)\\ \Leftrightarrow x^2+4x+4-9x^2+32x-32=0\\ \Leftrightarrow-8x^2+36x-28=0\\ \Leftrightarrow-2x^2+9x-7=0\\ \Leftrightarrow-7+7x+2x-2x^2=0\\ \Leftrightarrow-7\cdot\left(1-x\right)+2x\cdot\left(1-x\right)=0\\ \Leftrightarrow\left(-7+2x\right)\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-7+2x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{7}{2}\\x=1\end{matrix}\right.\)

a) 4         b) 0,6 hoặc 1,75     e) -3,5 hoặc -3