- Chọn D

⇔ 5√x - 4√x = 9 ⇔ √x = 9 ⇔ x = 81

Chị thử từng cái :)

là đáp án D. 81 nha

D nha bn !

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

\(\sqrt{25x}-\sqrt{16x}=9\) khi \(x\) bằng :

(A) 1 (B) 3 (C) 9 (D) 81

ĐK: x \(\ge\) 0

Ta có: \(\sqrt{25x}-\sqrt{16x}=9\Leftrightarrow5\sqrt{x}-4\sqrt{x}=9\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\left(tm\right)\)

Vậy đáp án đúng là D

ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}-\sqrt{16\left(x+1\right)}+\sqrt{9\left(x+1\right)}-\sqrt{4\left(x+1\right)}+\sqrt{x+1}=27\)

\(\Leftrightarrow5\sqrt{x+1}-4\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}+\sqrt{x+1}=27\)

\(\Leftrightarrow3\sqrt{x+1}=27\)

\(\Leftrightarrow\sqrt{x+1}=9\)

\(\Rightarrow x+1=81\)

\(\Rightarrow x=80\) (thỏa mãn)

a) Với x = 24

=> x + 1 = 24 (1)

Thay (1) vào A ta được:

\(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(A=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+x+1\)

\(A=1\)

b) Với x = 31

=> x - 1 = 30 (1)

Thay (1) vào B ta được

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

\(B=x^3-x^3+x^2-x^2+x+1\)

\(B=x+1\)

\(B=31+1=32\)

c) Với x = 14

=> x + 1 = 15

x + 2 = 16

2x + 1 = 29

x - 1 = 13

Thay tất cả biểu thức trên vào C ta được

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

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

\(C=-x\)

\(C=-14\)

d) Ta có:

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

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

\(\Rightarrow-2+x^2=1\)

\(\Rightarrow x^2=1+2=3\)

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

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

\(\sqrt{16x+16}+\sqrt{9x+9}-\sqrt{25x+25}+2\sqrt{x+1}=8\)

\(\Rightarrow4\sqrt{x+1}+3\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=8\)

\(\Rightarrow\sqrt{x+1}\left(4+3-5+2\right)=8\)

\(\Rightarrow4\sqrt{x+1}=8\)

\(\Rightarrow\sqrt{x+1}=2\)

\(\Rightarrow x+1=4\)

\(\Rightarrow\)\(x=3\)

\(\sqrt{16x+16}\) + \(\sqrt{9x+9}\) - \(\sqrt{25x+25}\) + 2\(\sqrt{x+1}\) = 8 ( x\(\ge\) -1)

<=> 4\(\sqrt{x+1}\) + 3\(\sqrt{x+1}\) - 5\(\sqrt{x+1}\) + 2\(\sqrt{x+1}\) = 8

<=> 4\(\sqrt{x+1}\) = 8

<=> \(\sqrt{x+1}\) = 2

<=> x + 1 =4

<=> x=3 (TM)

ta có:

\(\sqrt{16x+16}+\sqrt{9x+9}-\sqrt{25x+25}+2\sqrt{x+1}=8\)

=> \(4\cdot\sqrt{x+1}+3\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=8\)

=>\(\sqrt{x+1}\cdot\left(4+3-5+2\right)=8\)

=>\(4\sqrt{x+1}=8\)

=> \(\sqrt{x+1}=2\)

=>x+1=3

=> x=2