\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)

\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)

\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)

=>\(1.5\cdot\sqrt{x-1}=6\)

=>\(\sqrt{x-1}=4\)

=>x-1=16

=>x=17

\(\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\) (1)

\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{25}\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}=12+2\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-2\sqrt{x-1}=12\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

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

\(\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=16+1\)

\(\Leftrightarrow x=17\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{17\right\}\)

A) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\)

\(\Leftrightarrow\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=17\)

Vậy, x=17

A: \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

=>5/2*căn x-1-căn x-1=6

=>3/2*căn x-1=6

=>căn x-1=4

=>x-1=16

=>x=17

B:

a: ĐKXĐ: x>=0; x<>1

b: Sửa đề: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)

=căn x-1+x-căn x+1

=x

B) a) \(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b)Sửa đề \(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}-1+x-\sqrt{x}+1=x\)

6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\) 

\(\Leftrightarrow x^2-4x+1=x^2\)

\(\Leftrightarrow x^2-x^2=4x-1\)

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\) 

8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\) 

\(\Leftrightarrow x^2-x-6=x-3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1\)

\(\Leftrightarrow x=2\left(tm\right)\)

a. \(\Rightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\Rightarrow\sqrt{x+5}\left(2-3+4\right)=6\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)

b.\(\Rightarrow5\sqrt{x-1}-\frac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\Rightarrow\sqrt{x-1}\left(5-\frac{5}{2}-1\right)=6\Rightarrow\sqrt{x-1}=4\Rightarrow x-1=16\Rightarrow x=17\)

Mình làm ý a thôi nha còn ý b tương tự 

a, A = x⁵ - 5x⁴ + 5x³ - 5x² + 5x - 1

A= x⁵ - ( 4+1 ) x⁴ + ( 4+1 ) x³ - ( 4+1) x² + ( 4+1 ) x -1

Thay 4 = x vào biểu thức A, ta đc :

A = x⁵ - ( x+1 ) x⁴ + ( x+1 ) x³ - ( x+1 ) x² + ( x+1 ) x - 1

A = x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x -1

A = x -1

Thay x = 4 vào biểu thức A, ta đc :

A = 4 -1 

A = 3

b, B = x⁷ - 80x⁶ + 80x⁵ - 80x⁴ + .....+ 80x + 15

B = x⁷ - ( 79 +1 ) x⁶ + ( 79+1 )x⁵ - ( 79+1 ) x⁴ +....+( 79+1 )x + 15

Thay 79 = z vào biểu thức A, ta có :

B = x⁷ - ( x + 1 )x⁶ + ( x+1 )x⁵ - ( x+1 )x⁴ + .....+ ( x+1 )x +15

B= x⁷ - x⁷ - x⁶ + x⁶ + x⁵ - x⁵ - x⁴ + .....- x² + x² + x + 15

B= x + 15

Thay x= 79 vào biểu thức A, ta có:

A = 79 + 15

A= 94

c, C = x¹⁴ - 10x¹³ + 10x¹² - 10x¹¹ + ....+ 10x² - 10x + 10

C= x¹⁴ - ( x +1 )x¹³ + ( x + 1 ) x¹² - ( x + 1 )x¹¹ + ..... + ( x + 1 )x² - ( x + 1 )x - 10

C= x¹⁴ - x¹⁴ - x¹³ + x¹³ + x¹² - x¹² - x¹¹ +....+ x³ - x² + x² - x +10

C= -x -10 

Thay -x = -9 vào biểu thức C, ta có :

C = -9 + 10

C = 1

d, D = x¹⁰ - ( x+1 )x⁹ + (x + 1 )x⁸ - ( x+1 )x⁷ +....+( x+1 )x² - ( x + 1 )x + 25

D = x¹⁰ - ( x + 1 ) x⁹ + ( x + 1 )x⁸ - ( x + 1 )x⁷ + ..... + x³ - x² + x² - x + 25

D = -x + 25

thay -x = -24, vào biểu thức A , ta đc ;

A = -24 + 25

A = 1

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x+3\)

\(=3\)

Ta có : 

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)\(A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+3\)

\(A=3\)

P/s tham khảo nha 

hok tốt