\(4\sqrt{x+3}-\sqrt{x-1}=7\left(1\right)\left(ĐK:x\ge1\right)\)

\(\Leftrightarrow4\sqrt{x+3}=7+\sqrt{x-1}\)

\(\Leftrightarrow16\left(x+3\right)=49+x-1+14\sqrt{x+1}\)

\(\Leftrightarrow15x=14\sqrt{x-1}\)\(\left(x\ge1\right)\)

\(\Leftrightarrow225x^2=196\left(x-1\right)\)

\(\Leftrightarrow225x^2-196x+196=0\)

\(\Delta=196^2-4.225.196< 0\)

\(\Rightarrow pt\)vô nghiệm

Vậy pt vô nghiệm.

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

Đặt \(\sqrt{3x-1}=a\ge0\)

\(\Rightarrow\hept{\begin{cases}3x=a^2+1\\3x+2=a^2+3\\3x-1=a^2\end{cases}}\)

Pt (1) \(\Leftrightarrow3x\left(3x-1\right)-\left(3x+2\right)\sqrt{3x-1}+2=0\)

\(\Leftrightarrow\left(a^2+1\right)a^2-\left(a^2+3\right)a+2=0\)

\(\Leftrightarrow a^3+a^2-a^3-3a+2=0\)

\(\Leftrightarrow a^2-3a+2=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}}\)

TH1: a=1 

\(\Rightarrow\sqrt{3x-1}=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow x=\frac{2}{3}\left(tm\right)\)

TH2: a=2 

\(\Rightarrow\sqrt{3x-1}=2\)

\(\Leftrightarrow3x-1=4\)

\(\Leftrightarrow x=\frac{5}{3}\left(tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{\frac{2}{3};\frac{5}{3}\right\}\)

ĐK: \(x\ge1\).

Đặt \(\sqrt{x-1}=a,2x-5=b\).

Phương trình ban đầu tương đương với: 

\(\sqrt{b^2+5a^2}=2b+7a\)

\(\Rightarrow b^2+5a^2=4b^2+49a^2+28ab\)

\(\Leftrightarrow\left(2a+b\right)\left(22a+3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2a=-b\\22a=-3b\end{cases}}\)

Với \(2a=-b\Rightarrow2\sqrt{x-1}=5-2x\)

\(\Rightarrow4\left(x-1\right)=25-20x+4x^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=3-\frac{\sqrt{7}}{2}\\x=3+\frac{\sqrt{7}}{2}\end{cases}}\)

Thử lại chỉ có \(x=3-\frac{\sqrt{7}}{2}\)thỏa mãn. 

Với \(22a=-3b\Rightarrow22\sqrt{x-1}=-3\left(2x-5\right)\)

\(\Rightarrow484\left(x-1\right)=9\left(2x-5\right)^2\)

\(\Leftrightarrow x=\frac{83}{9}\pm\frac{55\sqrt{7}}{18}\)

Thử lại đều không thỏa mãn.

ĐK: x \(\ge\)-4

Ta có: \(x^2+\sqrt{x+4}+\sqrt{x+1}=x+27\)

<=> \(x^2-x-27+\sqrt{x+4}+\sqrt{x+11}=0\)

<=> \(x^2-x-20+\sqrt{x+4}-3+\sqrt{x+11}-4=0\)

<=> \(\left(x-5\right)\left(x+4\right)+\frac{x+4-9}{\sqrt{x+4}+3}+\frac{x+11-16}{\sqrt{x+4}+4}=0\)

<=> \(\left(x-5\right)\left(x+4\right)+\frac{x-5}{\sqrt{x+4}+3}+\frac{x-5}{\sqrt{x+4}+4}=0\)

<=> \(\left(x-5\right)\left(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}\right)=0\)

<=> \(x=5\)

(vì x \(\ge\)-4 => \(x+4\ge0\); \(\frac{1}{\sqrt{x+4}+3}>0\); \(\frac{1}{\sqrt{x+11}+4}>0\) 

=> \(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}>0\))

Đk: \(-2\le x\le3\)

Ta có: \(\sqrt{x+2}-\sqrt{3-x}=x^2-6x+9\)

<=> \(\sqrt{x+2}-2-\sqrt{3-x}+1-x^2+6x-8=0\)

<=> \(\frac{x+2-4}{\sqrt{x+2}+2}-\frac{3-x-1}{\sqrt{3-x}+1}-\left(x-4\right)\left(x-2\right)=0\)

<=> \(\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{3-x}+1}-\left(x-4\right)\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4=0\left(1\right)\end{cases}}\)

Vì \(-2\le x\le3\)

Do đó: \(\frac{1}{\sqrt{x+2}+2}>0\); \(\frac{1}{\sqrt{3-x}+1}>0\); 4 - x > 0

=> \(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4>0\)

=> pt (1) vô nghiệm

Vậy S = {2}

a. ta có 

\(A=\frac{\sqrt{20}-3\sqrt{4}}{\sqrt{14-6\sqrt{5}}}-\frac{\sqrt{20}-\sqrt{28}}{\sqrt{12-2\sqrt{35}}}\)\(\left(\text{ Nhân cả tử và mẫu với }\sqrt{2}\right)\)

\(=\frac{2\sqrt{5}-6}{\sqrt{\left(3-\sqrt{5}\right)^2}}-\frac{2\sqrt{5}-2\sqrt{7}}{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}=\frac{2\sqrt{5}-6}{3-\sqrt{5}}-\frac{2\sqrt{5}-2\sqrt{7}}{\sqrt{7}-\sqrt{5}}\)

\(=-2+2=0\)

b. \(A=\sqrt{\frac{\left(9-4\sqrt{3}\right)\left(6-\sqrt{3}\right)}{36-3}}-\sqrt{\frac{\left(3+4\sqrt{3}\right)\left(5\sqrt{3}+6\right)}{25\times3-36}}\)

\(A=\sqrt{\frac{66-33\sqrt{3}}{33}}-\sqrt{\frac{78+39\sqrt{3}}{39}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

ta có A<0 và \(A^2=2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}=2\)

Vậy \(A=-\sqrt{2}\)

ĐKXĐ : \(-2\le x\le1\)

Ta có : \(\sqrt{1-x}-\sqrt{2+x}=1\)

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

\(\Leftrightarrow1-x=1+2+x+2\sqrt{2+x}\)

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

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

\(\Leftrightarrow2+x=x+1\)

\(\Leftrightarrow0x=1\)(Vô lí)

Vậy PT vô nghiệm

a. ĐKXĐ: \(-2\le x\le1\)

ta có :\(\sqrt{1-x}=1+\sqrt{2+x}\Leftrightarrow1-x=1+2\sqrt{2+x}+2+x\)

\(\Leftrightarrow-2-2x=2\sqrt{2+x}\Leftrightarrow-1-x=\sqrt{2+x}\Leftrightarrow\hept{\begin{cases}x\le-1\\x^2+2x+1=x+2\end{cases}\Leftrightarrow x=\frac{-1-\sqrt{5}}{2}}\)

b.ĐKXĐ: \(-4\le x\le1\)ta có :

\(\sqrt{1-x}+\sqrt{4+x}=3\Leftrightarrow1-x+2\sqrt{1-x}\sqrt{4+x}+4+x=9\)

\(\Leftrightarrow\sqrt{1-x}.\sqrt{4+x}=2\Leftrightarrow4-3x-x^2=4\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Áp dụng bất đẳng thức Bunhia ta có :

\(\left(\sqrt{1+x^2}+\sqrt{2x}\right)^2\le2\left(1+x^2+2x\right)=2\left(x+1\right)^2\text{ nên }\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)

tương tự ta có : \(\hept{\begin{cases}\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\\\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\end{cases}}\)

Nên \(A\le\sqrt{2}\left(x+y+z+3\right)+\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(2-\sqrt{2}\right)\)

\(\le6\sqrt{2}+\left(2-\sqrt{2}\right)\sqrt{3\left(x+y+z\right)}\le6\sqrt{2}+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)

dấu bằng xảy ra khi x=y=z=1

ủa bạn oi nó là \(\sqrt{2}x\)mà có phai\(\sqrt{2x}dau\)