ĐK :\(x\ge\frac{2}{3}\)

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

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

Đặt \(x=\sqrt{3x-2}\) ; \(y=x+1\), pt trở thành :

\(x-y=\left(x^2-y^2\right)y^2\)

\(\Leftrightarrow\left(x-y\right)-y^2\left(x-y\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[y^2\left(x+y\right)-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\y^2\left(x+y\right)-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy^2+y^3-1=0\end{cases}}\)

Đến đây dễ rồi bạn tự giải tiếp nhé, xét các trường hợp là ra

ĐKXĐ : \(x\ne4\)

\(A=\frac{3}{2}\)

\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{3}{2}=0\)

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

\(\Leftrightarrow\frac{2\sqrt{x}+4-3\sqrt{x}+6}{2\left(\sqrt{x}-2\right)}=0\)

\(\Leftrightarrow\frac{10-\sqrt{x}}{2\left(\sqrt{x}-2\right)}=0\)

Mà \(2\left(\sqrt{x}-2\right)>0\)với \(x\ne4\)

\(\Rightarrow10-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=10\)

\(\Leftrightarrow x=100\)( thỏa mãn ĐKXĐ )

Vậy ..............

a) \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)  (đk: \(x\ge0\))

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

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

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

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

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

\(\Leftrightarrow x=9\)(tmđk)

vậy nghiệm của phtrinh là x = 9

b) \(\sqrt{x^2-6x+9}=6\)     (đk: \(x^2-6x+9\ge0\))

bình phương 2 vế, ta được: \(x^2-6x+9=36\)

\(\Leftrightarrow x^2-6x-27=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+3\right)=0\)

\(\Leftrightarrow x=9\)hoặc \(x=-3\)

a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{4+4\sqrt{3}+3}\)

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

\(=2-\sqrt{3}+2+\sqrt{3}=4\)

b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\left(a-2\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)

a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)

\(=\left|2-\sqrt{3}\right|+\sqrt{3+4\sqrt{3}+4}\)

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

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

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

\(=4\)

b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)( \(\hept{\begin{cases}a,b\ge0\\a\ne b\end{cases}}\))

\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(a-2\sqrt{ab}+b\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a-2\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a-b}{a-b}=1\)

\(a^2+b=b^2+c=c^2+a\)

\(\Leftrightarrow\hept{\begin{cases}a^2+b-b^2-c=0\\b^2+c-c^2-a=0\\c^2+a-a^2-b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=c-b\\b^2-c^2=a-c\\c^2-a^2=b-a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b\right)=c-b\\\left(b-c\right)\left(b+c\right)=a-c\\\left(c-a\right)\left(c+a\right)=b-a\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a+b=\frac{c-b}{a-b}\\b+c=\frac{a-c}{b-c}\\c+a=\frac{b-a}{c-a}\end{cases}}\)

 \(\Rightarrow\hept{\begin{cases}a+b-1=\frac{c-a}{a-b}\\b+c-1=\frac{a-b}{b-c}\\c+a-1=\frac{b-c}{c-a}\end{cases}}\)( * )

Thay ( * ) vào T ta được : \(T=\frac{\left(c-a\right)\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

Vậy T = 1

với x\(\inℤ\)

Sai đề. Phản ví dụ : z=y=-1, x =6

Nếu đề bài cho x, y, z cùng dương, ta làm như sau:

x+y >= 2 căn (xy)

\(\Rightarrow x+y \geq xyz \,\, nếu \\ 2 \geq z. \sqrt{xy}\)

Dùng bất đẳng thức cosi: x + y+ z/2 +z/2 >= ....

Sr tại mk ghi sai đề :))

ta có: \(a^3-3a^2+8a=9\)

\(\Leftrightarrow a^3-3a^2+8a-9=0\)

\(\Leftrightarrow a^3-3a^2+3a-1+5a-8=0\)

\(\Leftrightarrow\left(a-1\right)^3+5a-8=0\)(1)

và \(b^3-6b^2+17b=15\)biến đổi tương tự như a, ta được: \(\left(b-2\right)^3+5b-7=0\)(2)

Lấy (1) + (2) vế theo vế, ta được: \(\left(a-1\right)^3+\left(b-2\right)^3+5a-8+5a-7=0\)

\(\Leftrightarrow\left(a-1\right)^3+\left(b-2\right)^3+5\left(a+b-3\right)=0\)(3)

áp dụng hằng đẳng thức \(A^3+B^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)với \(A=a-1\)và \(B=b-2\)

ta được (3) <=> \(\left(a+b-3\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2\right]+5\left(a+b-3\right)=0\)

\(\Leftrightarrow\left(a+b-3\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\right]=0\)

vì \(\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\right]\ne0\)

\(\Rightarrow a+b-3=0\Rightarrow a+b=3\)