1. \(4x^2-49=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

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2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

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3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

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

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

hay x=6

\(a,\left(-5\right).\left|x\right|=-75\)

\(\left|x\right|=\frac{-75}{-5}=15\)

\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

Vậy....

\(b,\left(-6\right)^3.x^2=-1944\)

\(-216.x^2=-1944\)

\(x^2=9\)

\(\Rightarrow x=\pm3\)

Vậy....

\(d,\left|9-x\right|=-7+64\)

\(\left|9-x\right|=57\)

\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)

Vậy...

\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)

\(\left|x+101\right|+16=215\)

\(\left|x+101\right|=199\)

\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)

Vậy..

hok tốt!!

a,\(\left(-5\right).\left|x\right|=-75\)

\(=>\left|x\right|=-75:\left(-5\right)=15\)

\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

b,\(\left(-6\right)^3.x^2=-1944\)

\(=>\frac{1944}{216}=x^2\)

\(=>x=\sqrt{\frac{1944}{216}}=3\)

CẢM ƠN MỌI NGƯỜI NHIỀU Ạ 

Bài 9,

6²x73+36x3³=36x73+36x27=36(73+27)=36x100=3600.

197-\([\)6x(5-1)²+2022⁰\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.

Bài 10,

21-4x=13

=>4x=21-13=8

=>x=8:4=2.

30:(x-3)+1=4⁵:4³=4²=16

=>30:(x-3)=16-1=15

=>x-3=30:15=2

=>x=2+3=5.

(x-1)³+5x6=38

=>(x-1)³+30=38

=>(x-1)³=38-30=8=2³

=>x-1=2

=>x=3.

\(x^3-4x^2-9x+36=0\)

=> \(x^2\left(x-4\right)-9\left(x-4\right)=0\)

=> \(\left(x-4\right)\left(x^2-9\right)=0\)

=> \(\orbr{\begin{cases}x-4=0\\x^2-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\pm3\end{cases}}\)

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

=> \(\left(x^2-9+x-3\right)\left[x^2-9-\left(x-3\right)\right]=0\)

=> \(\left(x^2+x-12\right)\left(x^2-9-x+3\right)=0\)

=> \(\left(x^2+x-12\right)\left(x^2-x-6\right)=0\)

=> \(\left(x^2-3x+4x-12\right)\left(x^2+2x-3x-6\right)=0\)

=> \(\left[x\left(x-3\right)+4\left(x-3\right)\right]\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)

=> \(\left(x-3\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)=0\)

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

=> \(\hept{\begin{cases}\left(x-3\right)^2=0\\x+4=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-4\\x=-2\end{cases}}\)

\(x^3-3x+2=0\)

=> \(x^3-x-2x+2=0\)

=> \(x^2\left(x-1\right)-2\left(x-1\right)=0\)

=> \(\left(x-1\right)\left(x^2-2\right)=0\)

=> x = 1

dài thế

dai the giai bao gio xong chac mua quyt nam sau moi giai xong

a)\(\left(5x-1\right)^2-196=0\)

\(\Leftrightarrow\left(5x-1\right)^2=196\)

\(\Leftrightarrow5x-1=14\)

\(\Leftrightarrow x=3\)

b)\(4x^2+\frac{1}{4}=2x\)

\(\Leftrightarrow4x^2+\frac{1}{4}-2x=0\)

\(\Leftrightarrow\left(2x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow2x+\frac{1}{2}=0\)

\(\Leftrightarrow x=-\frac{1}{4}\)

c)\(x^2-12x=-36\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

\(\Leftrightarrow x=6\)

#H

a) (5x - 1)² - 196 = 0

<=> (5x - 1 - 14)(5x - 1 + 14) = 0

<=> (5x - 15)(5x + 13) = 0

<=> \(\orbr{\begin{cases}5x-15=0\\5x+13=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=3\\x=-\frac{13}{5}\end{cases}}\)

Vậy S = {3; -13/5}

b) Ta có: 4x² + 1/4 = 2x

<=> 16x² - 8x + 1  = 0

<=> (4x - 1)² = 0

<=> 4x-  1 = 0

<=> x = 1/4

Vậy S = {1/4}

c) x² - 12x = -36

<=> x² - 12x + 36 = 0 

<=> (x - 6)² 0 

<=> x - 6 = 0

<=> x = 6

Vậy S = {6}

Trả lời:

a, ( 5x - 1 )² - 196 = 0

<=> ( 5x - 1 - 14 ) ( 5x - 1 + 14 ) = 0

<=> ( 5x - 15 ) ( 5x + 13 ) = 0

<=> 5x - 15 = 0 hoặc 5x + 13 = 0

<=> 5x = 15 hoặc 5x = - 13

<=> x = 3 hoặc x = - 13/5

Vậy x = 3; x = - 13/5 là nghiệm của pt.

b, 4x² + 1/4 = 2x

<=> 4x² - 2x + 1/4 = 0

<=> ( 2x )² - 2.2x.1/2 + 1/4 = 0

<=> ( 2x - 1/2 )² = 0

<=> 2x - 1/2 = 0

<=> 2x = 1/2

<=> x = 1/4

Vậy x = 1/4 là nghiệm của pt.

c, x² - 12x = - 36

<=> x² - 12x + 36 = 0

<=> x² - 2.x.6 + 6² = 0

<=> ( x - 6 )² = 0

<=> x - 6 = 0

<=> x = 6

Vậy x = 6 là nghiệm của pt.