kmjkjpjpi

mình quên ĐK là x,y,z dương

\(ĐK:x\ge-3\)

\(\left(2x-5\right)\left(\sqrt{x+3}-1\right)=2x^2-x-10\)

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

\(\Leftrightarrow\left(2x-5\right)\left(x+2-\sqrt{x+3}+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x-\sqrt{x+3}+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\x-\sqrt{x+3}+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\left(tm\right)\\\sqrt{x+3}=x+2\left(đk:x\ge-2\right)\end{cases}}}\)

+) \(\sqrt{x+3}=x+2\Leftrightarrow x+3=x^2+4x+4\)

\(\Leftrightarrow x^2+3x+1=0\) 

\(\Delta=b^2-4ac=3^2-4=5\) \(\Rightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\left(tm\right)\\x=\frac{-3-\sqrt{5}}{2}\left(loai\right)\end{cases}}\)

vậy_

\(\left(2x-5\right)\left(\sqrt{x+3}-1\right)=2x^2-x-10\)ĐK : x>= - 3 

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

\(\Leftrightarrow\left(2x-5\right)\left(\sqrt{x+3}-1-x-2\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(\sqrt{x+3}-3-x\right)=0\)

TH1 : \(2x-5=0\Leftrightarrow x=\frac{5}{2}\)(tm) 

TH2 : \(\sqrt{x+3}=x+3\Leftrightarrow x+3=x^2+6x+9\)

\(\Leftrightarrow x^2+5x+6=0\)

\(\Delta=25-24=1>0\)pt có 2 nghiệm phân biệt 

\(x_1=\frac{-5-1}{2}=-3;x_2=\frac{-5+1}{2}=-2\)(tm)

\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)

\(=6\cdot\frac{\sqrt{3}}{2}+10\cdot\frac{2\sqrt{3}}{5}-15\cdot\frac{4}{\sqrt{3}}+9\cdot\frac{2}{\sqrt{3}}\)

\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}=-7\sqrt{3}\)

Trả lời:

\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)

\(=6.\frac{\sqrt{3}}{\sqrt{4}}+10.\frac{\sqrt{12}}{\sqrt{25}}-15.\frac{\sqrt{16}}{\sqrt{3}}+9.\frac{\sqrt{4}}{\sqrt{3}}\)

\(=6.\frac{\sqrt{3}}{2}+10.\frac{\sqrt{2^2.3}}{5}-15.\frac{4}{\sqrt{3}}+9.\frac{2}{\sqrt{3}}\)

\(=3\sqrt{3}+10.\frac{2\sqrt{3}}{5}-15.\frac{4\sqrt{3}}{3}+9.\frac{2\sqrt{3}}{3}\)

\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}\)

\(=\left(3+4-20+6\right).\sqrt{3}=-7\sqrt{3}\)

|x + 2| + |x - 4| = 0

có |x + 2| ≥ 0 và |x - 4| ≥ 0

=> x + 2 = 0 và x - 4 = 0

=> x = -2 và x = 4 (vô lí)

vậy pt vô nghiệm

Với \(x\ge0;x\ne1;\frac{1}{4}\)

\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\)

\(=\left(\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\left(\frac{\sqrt{x}+\sqrt{x}+3}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{2\sqrt{x}-1}=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)

đề bài là rút gon biểu thức đấy 

A, \(P=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

\(P=\frac{x-1}{\sqrt{x}}:\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b, \(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{4-2\sqrt{3}}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

thay vào P ta được 

\(P=\frac{\left(\sqrt{\left(\sqrt{3}+1\right)^2}+1\right)^2}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{\left(\left|\sqrt{3}+1\right|+1\right)^2}{\left|\sqrt{3}+1\right|}=\frac{\left(\sqrt{3}+2\right)^2}{\sqrt{3}+1}=\frac{5+4\sqrt{3}}{\sqrt{3}+1}\)

c, \(P-2=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-2=\frac{x+2\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}}=\frac{x+1}{\sqrt{x}}\) mà có x + 1 > 0 và căn x > 0

=> p - 2 > 0

=> p > 2

d, p.căn x = 6.căn x - 3 - căn (x - 4)

 thay p vào ta đc :

(căn x + 1)^2 = 6.căn x - 3 - căn (x-4)

<=> x + 2.căn x + 1 = 6.căn x - 3 - căn (x - 4)

<=> x - 4.căn x + 4 - căn.(x - 4) = 0

<=> (căn x - 2)^2 + căn (x - 4) = 0

có (căn x - 2)^2 ≥ 0 và căn (x - 4) ≥ 0

=> căn x - 2 = 0 và x - 4 = 0

=> căn x = 2 và x = 4

=> x = 4 (thỏa mãn)

vậy  

may tên Uyên mí giải cho đấy :))