ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-1}=a\ge0\\\sqrt[3]{2-x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^3=1\)

Ta được hệ: 

\(\left\{{}\begin{matrix}a+b=1\\a^2+b^3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+b^3=1\end{matrix}\right.\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x-1}=0\\\sqrt[]{x-1}=1\\\sqrt[]{x-1}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=10\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow1}\dfrac{2-\sqrt[]{2x-1}\sqrt[3]{5x+3}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{2-2\sqrt[]{2x-1}+2\sqrt[]{2x-1}-\sqrt[]{2x-1}.\sqrt[3]{5x+3}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{2\left(1-\sqrt[]{2x-1}\right)+\sqrt[]{2x-1}\left(2-\sqrt[3]{5x+3}\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{4\left(x-1\right)}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}\left(x-1\right)}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(-\dfrac{4}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}\right)\)

\(=-\dfrac{4}{1+1}-\dfrac{5\sqrt[]{1}}{4+4+4}=-\dfrac{29}{12}\)

ĐKXĐ: x>=1/5

TA có: \(\sqrt{5x-1}-\sqrt{x+3}=9\)

=>\(\left(\sqrt{5x-1}-\sqrt{x+3}\right)^2=9^2=81\)

=>\(5x-1+x+3-2\cdot\sqrt{\left(5x-1\right)\left(x+3\right)}=81\)

=>\(6x-2-2\sqrt{\left(5x-1\right)\left(x+3\right)}=81\)

=>\(\sqrt{4\left(5x-1\right)\left(x+3\right)}=6x-2-81=6x-83\)

=>\(\begin{cases}6x-83\ge0\\ \left(6x-83\right)^2=4\left(5x-1\right)\left(x+3\right)\end{cases}\Rightarrow\begin{cases}x>=\frac{83}{6}\\ 36x^2-996x+6889=4\left(5x^2+15x-x-3\right)\end{cases}\)

=>\(\begin{cases}x\ge\frac{83}{6}\\ 36x^2-996x+6889-20x^2-56x+12=0\end{cases}\Rightarrow\begin{cases}x\ge\frac{83}{6}\\ 16x^2-1052x+6901=0\end{cases}\)

\(16x^2-1052x\) +6901=0

=>\(\left(4x\right)^2-2\cdot4x\cdot131,5+17292,25=10391,25\)

=>\(\left(4x-131,5\right)^2=\frac{41565}{4}=\left(\frac{\sqrt{41565}}{2}\right)^2\)

=>\(\left[\begin{array}{l}4x-\frac{263}{2}=\frac{\sqrt{41565}}{2}\\ 4x-\frac{263}{2}=-\frac{\sqrt{41565}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=\frac{263+\sqrt{41565}}{2}\\ 4x=\frac{263-\sqrt{41565}}{2}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{263+\sqrt{41565}}{8}\\ x=\frac{263-\sqrt{41565}}{8}\end{array}\right.\)

mà x>=83/6

nên x=\(\frac{263+\sqrt{41565}}{8}\)

ĐKXĐ: x > -3 

            y > -1

Đặt \(\hept{\begin{cases}\sqrt{x+3}=a\left(a\ge0\right)\\\sqrt{y+1}=b\left(b\ge0\right)\end{cases}}\) thì hệ đã cho trở thành

\(\hept{\begin{cases}2a-3b=2\\a-b=1\end{cases}\Leftrightarrow}\hept{\begin{cases}2a-3b=2\\2a-2b=2\end{cases}\Leftrightarrow}\hept{\begin{cases}b=0\\a=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+3}=1\\\sqrt{y+1}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=-1\end{cases}\left(tm\right)}\)

a: ĐKXĐ: -21<=x<=21 và x<>0

Ta có: \(\frac{\sqrt{21+x} + \sqrt{21-x}}{\sqrt{21+x} - \sqrt{21-x}} = \frac{21}{x}\)

=>\(\frac{(\sqrt{21+x} + \sqrt{21-x})^2}{(21+x) - (21-x)} = \frac{21}{x}\)

=>\(\frac{(21+x) + (21-x) + 2\sqrt{(21+x)(21-x)}}{2x} = \frac{21}{x}\)

=>\(\frac{42 + 2\sqrt{441 - x^2}}{2x} = \frac{21}{x}\)

=>\(\frac{21 + \sqrt{441 - x^2}}{x} = \frac{21}{x}\)

=>\(21+\sqrt{441-x^2}=21\)

=>\(\sqrt{441-x^2}=0\)

=>\(441-x^2=0\)

=>\(x^2=441\)

=>x=21(nhận) hoặc x=-21(nhận)

b: ĐKXĐ: x∈R

\(\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)^3\)

\(=x+1+3x+1+3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

=4x+2+\(3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

Phương trình ban đầu sẽ trở thành:

\(4x + 2 + 3\sqrt[3]{(x+1)(3x+1)(x-1)} = x - 1\)

=>\(3\sqrt[3]{(x^2-1)(3x+1)}=-3x-3\)

=>\(\sqrt[3]{(x^2-1)(3x+1)}=-(x+1)\)

=>\((x^2-1)(3x+1) = -(x+1)^3\)

=>\((x-1)(x+1)(3x+1) + (x+1)^3 = 0\)

=>(x+1)(3x^2+x-3x-1+x^2+2x+1)=0

=>(x+1)(4x^2)=0

=>x=0 hoặc x=-1

Khi x=0 thì \(\sqrt[3]{1}+\sqrt[3]{1}=2<>\sqrt[3]{-1}=-1\) (loại)

Khi x=-1 thì \(\sqrt[3]{0}+\sqrt[3]{-2}=\sqrt[3]{-2}\) (nhận)

a: ĐKXĐ: 3<=x<=8

\(\sqrt{\left(x-3\right)\left(8-x\right)}+26>-x^2+11x\)

=>\(\sqrt{8x-x^2-24+3x}+26>-x^2+11x\)

=>\(\sqrt{-x^2+11x-24}>-x^2+11x-26\) (1)

Đặt \(a=\sqrt{-x^2+11x-24}\) (ĐIều kiện: a>=0)

=>\(a^2=-x^2+11x-24\)

=>\(a^2-2=-x^2+11x-24-2=-x^2+11x-26\)

(1) sẽ trở thành: \(a>a^2-2\)

=>\(a^2-2

=>\(a^2-a-2<0\)

=>(a-2)(a+1)<0

=>a-2<0

=>a<2

=>\(-x^2+11x-24<4\)

=>\(-x^2+11x-28<0\)

=>\(x^2-11x+28>0\)

=>(x-7)(x-4)>0

=>x>7 hoặc x<4

kết hợp ĐKXĐ, ta được: 3<=x<4 hoặc 7<x<=8

b: ĐKXĐ: x∈R

\(\left(x+1\right)\left(x+4\right)<5\cdot\sqrt{x^2+5x+28}\)

=>\(x^2+5x+4<5\cdot\sqrt{x^2+5x+28}\) (2)

Đặt \(a=\sqrt{x^2+5x+28}\) (Điều kiện: a>0)

=>\(a^2=x^2+5x+28\)

=>\(a^2-24=x^2+5x+4\)

(2) sẽ trở thành: \(a^2-24<5a\)

=>\(a^2-5a-24<0\)

=>(a-8)(a+3)<0

=>a-8<0

=>a<8

=>\(x^2+5x+28<64\)

=>\(x^2+5x-36<0\)

=>(x+9)(x-4)<0

=>-9<x<4

CÁi  này easy mà .-.

\(\frac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)

\(\Leftrightarrow\frac{\frac{\left(7-x\right)-\left(x-5\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)

\(\Leftrightarrow\frac{\frac{-2\left(x-6\right)}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(\frac{\frac{-2}{\left(\sqrt[3]{7-x}\right)^2+\left(\sqrt[3]{x-5}\right)^2+\sqrt[3]{7-x}\sqrt[3]{x-5}}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}+1\right)=0\)

\(\Rightarrow x-6=0\Rightarrow x=6\)

\(ĐK:x\ge1\)

\(PT\Leftrightarrow x+3-4\sqrt{x+3}+4+\sqrt{x-1}=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+3}=2\\x-1=0\end{cases}\Leftrightarrow}x=1\left(tm\right)\)

Lời giải:

Đặt $\sqrt[3]{x}=a; \sqrt[3]{2x-3}=b$. Ta có:

\(\left\{\begin{matrix} a+b=\sqrt[3]{4(a^3+b^3)}\\ 2a^3-b^3=3\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a^3+b^3+3ab(a+b)=4(a^3+b^3)\\ 2a^3-b^3=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^3+b^3=ab(a+b)\\ 2a^3-b^3=3\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} (a-b)^2(a+b)=0(1)\\ 2a^3-b^3=3(2)\end{matrix}\right.\)

Từ $(1)$ suy ra $a=b$ hoặc $a=-b$.

Nếu $a=b$. Thay vào $(2)$ suy ra $a^3=b^3=3$

$\Leftrightarrow x=2x-3=3$ (thỏa mãn)

Nếu $a=-b$. Thay vào $(2)$ suy ra $a^3=1; b^3=-1$

$\Leftrightarrow x=1; 2x-3=-1$ (thỏa mãn)

Vậy $x=3$ hoặc $x=1$