\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)

\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}=9\)

\(\Leftrightarrow x+2=81\)

\(\Leftrightarrow x=79\)

\(b,x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)

Đặt: \(\hept{\begin{cases}\sqrt{x-1}=a\\\sqrt{7-x}=b\end{cases}}\)Ta được pt mới: \(a^2+2b=2a+ab\Leftrightarrow\left(a-2\right)\left(a-b\right)=0\)

cái thứ 1 nhân liên hợp đi 

sau đó nhân chéo lên vs vế phải

rồi rút gọn

bình lên

giải pt là đc

Lời giải: 
ĐKXĐ: $x\geq 0; x\neq 1$
a.

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

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

\(=\frac{2(x-2\sqrt{x}+1)}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2}{x+\sqrt{x}+1}\)

b.

Ta thấy với $x\geq 0 ; x\neq 1$ thì $x+\sqrt{x}+1\geq 1$

$\Rightarrow A=\frac{2}{x+\sqrt{x}+1}\leq 2$

Vậy $A$ đạt max bằng $2$ khi $x=0$

\(\hept{\begin{cases}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\left(1\right)\\x^3-8x-1=2\sqrt{y-2}\left(2\right)\end{cases}}\)

\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{y\left(12-x^2\right)}=12-x\sqrt{12-y}\)

\(\Leftrightarrow\left(\sqrt{y\left(12-x^2\right)}\right)^2=\left(12-x\sqrt{12-y}\right)^2\)

\(\Leftrightarrow x^2-2x\sqrt{12-y}+\left(12-y\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{12-y}\right)^2=0\)

\(\Leftrightarrow3-y=x^2-9\left(3\right)\)

Ta lại có:

\(\left(2\right)\Leftrightarrow\left(x^3-8x-3\right)=2\left(\sqrt{y-2}-1\right)\)

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

Thay (3) vào (4) ta được:

\(\left(x-3\right)\left(x^2+3x+1\right)+\frac{2\left(x^2-9\right)}{\sqrt{y-2}+1}=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}\)

Đặt \(\sqrt{x^2-x+1}=a"ĐK:a>0"\)

\(pt\Leftrightarrow\frac{"6^2+3x^4a""4-a^2"}{4"2+a"a^2}=a"2-a"\)

\(\Leftrightarrow"x^6+3x^4a""4-a^2"=4a^3"4-a^2"\)

\(\Leftrightarrow"4-a^2""x^6+3x^4a-4a^3"=0\)

TH1: \(4-a^2=0\Leftrightarrow\orbr{\begin{cases}a=-2\\a=2\end{cases}}\)

Với \(a=2,\sqrt{x^2-x+1}=2\Rightarrow x^2-x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{cases}}\)

TH2: \(x^6+3x^4a-4a^3=0\Rightarrow x^6-4x^4a-4x^2a^2+4x^2a^2-4a^3=0\)

\(\Leftrightarrow"x^2-a""x^4+4x^2a+4a^2"=0\Leftrightarrow"x^2-a""x^2+2a"^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=a\\x^2=-2a\end{cases}}\)

Với \(x^2=a\Rightarrow x^2=\sqrt{x^2-x+1}\)

P/s: Tham khảo thôi đừng có chép nguyên vào

Thay dấu ngoặc kép thành ngoặc đơn nha

\(\hept{\begin{cases}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\left(1\right)\\x^3-8x-1=2\sqrt{y-2}\left(2\right)\end{cases}}\)

\(ĐK:-2\sqrt{3}\le x\le2\sqrt{3},2\le y\le12\)

Ta có: \(\left(a-b\right)^2\ge0\forall a,b\inℝ\Leftrightarrow\frac{a^2+b^2}{2}\ge ab\forall a,b\inℝ\)

Do đó: \(\hept{\begin{cases}x\sqrt{12-y}\le\left|x\right|\sqrt{12-y}\le\frac{x^2+12-y}{2}\\\sqrt{y\left(12-x^2\right)}\le\frac{y+12-x^2}{2}\end{cases}}\)

\(\Rightarrow x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}\le\frac{\left(x^2+12-y\right)+\left(y+12-x^2\right)}{2}=12\)

Suy ra \(\left(1\right)\Leftrightarrow\hept{\begin{cases}x\ge0\\y=12-x^2\end{cases}}\)

Thay \(y=12-x^2\)vào (2), ta được: \(x^3-8x-1=2\sqrt{10-x^2}\)

\(\Leftrightarrow x^3-8x-3+2\left(1-\sqrt{10-x^2}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+1\right)+\frac{2\left(x^2-9\right)}{1+\sqrt{10-x^2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left[x^2+3x+1+\frac{2\left(x+3\right)}{1+\sqrt{10-x^2}}\right]=0\)

Mà ta dễ thấy được: \(x^2+3x+1+\frac{2\left(x+3\right)}{1+\sqrt{10-x^2}}>0\forall x\ge0\)nên \(x-3=0\Leftrightarrow x=3\left(tm\right)\Rightarrow y=12-3^2=3\)

Vậy hệ có 1 nghiệm duy nhất \(\left(x,y\right)=\left(3,3\right)\)

\(\dfrac{x+1}{39}+\dfrac{x+2}{38}+\dfrac{x+3}{37}=0\)

\(\Leftrightarrow\dfrac{x+1}{39}+1+\dfrac{x+2}{38}+1+\dfrac{x+3}{37}+1-3=0\)

\(\Leftrightarrow\dfrac{x+40}{39}+\dfrac{x+40}{38}+\dfrac{x+40}{37}=3\)

\(\Leftrightarrow\left(x+40\right)\left(\dfrac{1}{39}+\dfrac{1}{38}+\dfrac{1}{37}\right)=3\)

\(\Leftrightarrow\left(x+40\right).\dfrac{4331}{54834}=3\)

\(\Leftrightarrow x+40=\dfrac{164502}{4331}\)

\(\Leftrightarrow x=\dfrac{-8738}{4331}\)

-Vậy \(S=\left\{\dfrac{-8738}{4331}\right\}\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

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

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)

\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)

\(< =>\dfrac{x-5}{4}=4x-17\)

\(< =>x-5=16x-68\)

\(< =>15x=68-5=63\)

\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)

b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)

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

\(< =>2x+1+1+2\sqrt{2x+1}=4x\)

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

\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)

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

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)

\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)