Yêu cầu đề là gì thế bạn? Bạn cần viết rõ ra thì mọi người mới giúp được chứ????

\(\left(3-4x\right)^2=25=5^2\)

\(\Rightarrow3-4x=5\)

\(\Rightarrow4x=3-5=-2\Rightarrow x=-\frac{1}{2}\)

\(\left(2x-\frac{1}{4}\right)^2=16=4^2\)

\(\Rightarrow2x-\frac{1}{4}=4\Rightarrow2x=4+\frac{1}{4}=\frac{17}{4}\)

\(\Rightarrow x=\frac{17}{4}:2=\frac{17}{4}.\frac{1}{2}=\frac{17}{8}\)

Đề số 3 bị sai.

\(\left(2x+5\right)^2=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=-\frac{5}{2}\)

(3-4x)²=25

3-4x=5

4x=3-5

4x=-2

x=-2:4

x=-0,5

b)(2x-1/4²)=16

2x-1/4=4

2x=4+1/4

2x=4,25

x=2,125

c) cái này x ở đâu vậy bn

d) (2x+5)²=0

2x+5=0

2x=0+5

2x=5

x=5:2

x=5/2

Nhớ k cho mk nha

a)\(\left(3-4x\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}3-4x=5\\3-4x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}4x=3-5\\4x=3-\left(-5\right)\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-2\\4x=8\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=2\end{cases}}\)

b) \(\left(2x-\frac{1}{4}\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{4}=4\\2x-\frac{1}{4}=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=-4+\frac{1}{4}\\2x=4+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-\frac{15}{4}\\2x=\frac{17}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-15}{4}:2\\x=\frac{17}{4}:2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{8}\\x=\frac{17}{8}\end{cases}}\)

c)\(\left(4-x\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4-4\\x=4-\left(-4\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}\)

d) \(\left(2x+5\right)^2=0\)

\(\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=\frac{-5}{2}\)

4(3x²-4)-6(2x²-2x)=-16

12x²-16-12x²+12x=-16

12x²-12x²+12x=16-16

12x=0

=>x=0

\(4\left(3x^2-4\right)-6\left(2x^2-2x\right)=-16.\)

\(12x^2-16-12x^2+12x=-16\)

\(12x^2-12x^2+12x=16-16\)

\(12x=0\)

Học tốt

a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ( x + 4 )² - ( x + 1 )( x - 1 ) = 16

<=> x² + 8x + 16 - ( x² - 1 ) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = -1/8

b) ( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 49 ) = 0

<=> 5x² + 2x + 10 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> 2x = -255/2

c) ( x + 2 )( x² - 2x + 4 ) - x( x² + 2 ) = 15

<=> x³ + 2³ - x³ - 2x = 15

<=> 8 - 2x = 15

<=> 2x = -7

<=> x = -7/2

d) ( x + 3 )³ - x( 3x + 1 )² + ( 2x + 1 )( 4x² - 2x + 1 ) = 28

<=> x³ + 9x² + 27x + 27 - x( 9x² + 6x + 1 ) + [ ( 2x )³ + 1³ ] = 28

<=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

<=> 3x² + 26x + 28 = 28

<=> 3x² + 26x = 0

<=> x( 3x + 26 ) = 0

<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)

=>6xy-8y-9x+12=6xy-15y+2x-5 và 2y-6+16=3x+6

=>-9x-8y+12+15y-2x+5=0 và 3x+6-2y-10=0

=>-11x+7y=-17 và 3x-2y=4

=>x=6 và y=7

bài này khó ak nha

Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)

<=>1/x+2 -1/x+6 +1/x+6 -1/x+13

<=>1/x+2-1/x+13

=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)

<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)

<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)

<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)

<=> 2x+1=14

<=> 2x=14-1

<=> 2x=13

<=> x=13:2

<=> x=13/2

Vậy x=13/2

Chắc là vầy. Mk cug ko chắc nữa

\(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(A=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{5^{32}-1}{2}\)

\(A=\left|2x+4\right|+\left|2x+6\right|+\left|2x+8\right|\)

\(A=\left|2x+4\right|+\left|2x+8\right|+\left|2x+6\right|\)

\(A=\left|2x+4\right|+\left|-2x-8\right|+\left|2x+6\right|\)

\(A\ge\left|2x+4-2x-8\right|+\left|2x+6\right|\)

\(A\ge4+\left|2x+6\right|\)

Vì \(\left|2x+6\right|\ge0\) nên \(A\ge4\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x+4\le0\\2x+6=0\\2x+8\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\le-4\\2x=-6\\2x\ge-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\x=-3\\x\ge-4\end{matrix}\right.\)

Vậy \(x=-3\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)

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

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125