\(P=x^2+8x+16+x^2-25-2x^2-2x=6x-9\\ Q=y\left(x-4\right)-5\left(x-4\right)=\left(y-5\right)\left(x-4\right)\\ Q=\left(5,5-5\right)\left(14-4\right)=0,5\cdot10=5\)

Đáp án đúng : B

Chọn A

A

chon A bạn nhé

Ta có

A   =   5 ( x   +   4 ) 2   +   4 ( x   –   5 ) 2   –   9 ( 4   +   x ) ( x   –   4 )     =   5 ( x 2   +   2 . x . 4   +   16 )   +   4 ( x 2   –   2 . x . 5   +   5 2 )   –   9 ( x 2   –   4 2 )     =   5 ( x 2   +   8 x   +   16 )   +   4 ( x 2   –   10 x   +   25 )   –   9 ( x 2   –   4 2 )     =   5 x 2   +   40 x   +   80   +   4 x 2   –   40 x   +   100   –   9 x 2   +   144     = ( 5 x 2   +   4 x 2   –   9 x 2 )   +   ( 40 x   –   40 x )   +   ( 80   + 100   +   144 )

= 324

Đáp án cần chọn là: C

x + 22 + (-14) = x + [22 + (-14)]

= x + (22 -14) = x + 8

a, \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot3+3^2}\)

\(=\sqrt{5}-2+\sqrt{\left(\sqrt{5}-3\right)^2}\)

\(=\sqrt{5}-2+\left|\sqrt{5}-3\right|\)

\(=\sqrt{5}-2+3-\sqrt{5}\)

\(=1\)

b, (ĐKXĐ: x ≥ 0; x ≠ 1)

\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

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

\(=\dfrac{x-5}{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

#\(Toru\)

a: \(=\sqrt{5}-2+3-\sqrt{5}=3-2=1\)

b: 

ĐKXĐ: \(x\ge0,x\ne1\)

\(A=\dfrac{x-5+\sqrt{x}-1+2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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hình như là gpt :v

\(5\sqrt{x^2+9}+2\sqrt{x}=\frac{x^2}{2}+\frac{x}{2}+19\)

ĐK:\(x\ge0\)

\(pt\Leftrightarrow5\sqrt{x^2+9}-25+2\sqrt{x}-4=\frac{x^2}{2}+\frac{x}{2}-10\)

\(\Leftrightarrow\frac{25\left(x^2+9\right)-625}{5\sqrt{x^2+9}+25}+\frac{4x-16}{2\sqrt{x}+4}=\frac{x^2+x-20}{2}\)

\(\Leftrightarrow\frac{25x^2-400}{5\sqrt{x^2+9}+25}+\frac{4x-16}{2\sqrt{x}+4}-\frac{x^2+x-20}{2}=0\)

\(\Leftrightarrow\frac{25\left(x-4\right)\left(x+4\right)}{5\sqrt{x^2+9}+25}+\frac{4\left(x-4\right)}{2\sqrt{x}+4}-\frac{\left(x-4\right)\left(x+5\right)}{2}=0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{25\left(x+4\right)}{5\sqrt{x^2+9}+25}+\frac{4}{2\sqrt{x}+4}-\frac{\left(x+5\right)}{2}\right)=0\)

Suy ra x=4

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

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

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

\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\) (ĐK: \(x\ge0;x\ne1\)) 

\(A=\dfrac{\left(\sqrt{x}\right)^3+1^3}{\left(\sqrt{x}\right)^2-1^2}-\dfrac{\left(\sqrt{x}\right)^2-1^2}{\sqrt{x}+1}\)

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

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

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

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

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

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\) 

ĐK: \(\left\{{}\begin{matrix}x-2\sqrt{x}-3\ne0\\\sqrt{x}+1\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\ne0\\\sqrt{x}+1\ne0\left(hiển-nhiên\right)\\x\ne\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow x\ne\sqrt{3}\)

\(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt[]{x}-3}-\dfrac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)

\(\Leftrightarrow\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(-\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow\dfrac{x\sqrt{x}-3-2\left(x-9\right)-x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x+1}\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x-4\right)\sqrt{x}-3x+12}{\left(\sqrt{x+1}\right)\left(\sqrt{x}-3\right)}\)

Chúc bạn học tốt ^^

Sửa: \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)

\(A=\left(3x+2\right)^2-2\left(3x+2\right)\left(2x-7\right)+\left(2x-7\right)^2\)

\(A=\left[\left(3x+2\right)-\left(2x-7\right)\right]^2\)

\(A=\left(3x+2-2x+7\right)^2\)

\(A=\left(x+9\right)^2\)

Thay \(x=-19\) vào A ta có:

\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)

Vậy: ...