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

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

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

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

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

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

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

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

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

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

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

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

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

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 \(x=-3\)

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

\(\Rightarrow3-4x=5\)

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

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

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

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

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

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

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

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 ~~~ 

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

a, Ta có : \(\left(2x-1\right)^4=16\)

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

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

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

Mà \(\left(2x-1\right)^2+2^2>0\)

=> \(\left(2x-3\right)\left(2x+1\right)=0\)

=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2};-\frac{1}{2}\right\}\)

b, Ta có : \(\left(2x+1\right)^4=\left(2x+1\right)^6\)

=> \(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)

=> \(\left(2x+1\right)^4\left(\left(2x+1\right)^2-1\right)=0\)

=> \(\left(2x+1\right)^4\left(2x+1-1\right)\left(2x+1+1\right)=0\)

=> \(2x\left(2x+1\right)^4\left(2x+2\right)=0\)

=> \(\left[{}\begin{matrix}2x=0\\2x+1=0\\2x+2=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{0;-1;-\frac{1}{2}\right\}\)

c, Ta có : \(\left|\left|x+3\right|-8\right|=20\)

TH₁ : \(x+3\ge0\left(x\ge-3\right)\)

=> \(\left|x+3\right|=x+3\)

=> \(\left|x-5\right|=20\)

TH_1.1 : \(x-5\ge0\left(x\ge5\right)\)

=> \(\left|x-5\right|=x-5=20\)

=> \(x=25\left(TM\right)\)

TH_1.2 : \(x-5< 0\left(x< 5\right)\)

=> \(\left|x-5\right|=5-x=20\)

=> \(x=-15\) ( không thỏa mãn )

TH₂ : \(x+3< 0\left(x< -3\right)\)

=> \(\left|x+3\right|=-x-3\)

=> \(\left|-x-11\right|=20\)

TH_1.1 : \(-x-11\ge0\left(x\le-11\right)\)

=> \(\left|-x-11\right|=-x-11=20\)

=> \(x=-31\left(TM\right)\)

TH_1.2 : \(-x-11< 0\left(x>-11\right)\)

=> \(\left|-x-11\right|=x+11=20\)

=> \(x=9\) ( không thỏa mãn )

Vậy phương trình có tập nghiệm là \(S=\left\{-31;25\right\}\)

a, ( 2x - 1 )⁴ = 16

=> 2x - 1 = 2 hoặc -2

TH1: 2x - 1 = 2

=> 2x = 2 + 1 = 3; => x = \(\frac{3}{2}\)

TH2: 2x - 1 = -2

=> 2x = -2 + 1 = -1; => x =- \(\frac{1}{2}\)

b, ( 2x + 1 )⁴ = ( 2x + 1 )⁶

=> ( 2x + 1 )⁴ - ( 2x + 1 )⁶ = 0

= ( 2x + 1 )⁴ - ( 2x - 1 )² . ( 2x - 1 )⁴

= ( 2x + 1 )⁴ . [ 1 - ( 2x - 1 )² ] = 0

Ta có ( 2x + 1 )⁴ và ( 2x - 1 )² \(\ge\) 0 vì có số mũ chẵn

Ta có 2 TH

TH1: ( 2x - 1 )⁴ = 0

=> 2x - 1 = 0; => x = \(\frac{1}{2}\)

TH2: 1 - ( 2x - 1 )² = 0; => ( 2x - 1 )² = 1

=> 2x - 1 = 1; => x = 1

c, //x + 3/ - 8/ = 20

Ta có 2 TH, mỗi TH lại chia thành 2 TH nhỏ hơn

TH1: /x + 3/ - 8 = 20

=> /x + 3/ = 28

=> x + 3 = 28 hoặc -28

TH1 nhỏ: x + 3 = 28; => x = 25

TH2 nhỏ: x + 3 = -28; => x = -31

TH2: /x + 3/ - 8 = -20

=> /x + 3/ = -12; => TH này loại

=> x = 25; -31

=>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

Phương Văn Cảnh giúp mk bài này vs