Lập phương 2 vế lên bn

Giải dùm tui đi bạn

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

\(pt\Leftrightarrow\sqrt[3]{2x-3}-1+\sqrt[3]{x-2}=0\)

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

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

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

Dễ thấY :\(\frac{2}{\sqrt[3]{\left(2x-3\right)^2}+\sqrt[3]{2x-3}+1}+\frac{1}{\sqrt[3]{\left(x-2\right)^2}}>0\)

\(\Rightarrow x-2=0\Rightarrow x=2\). Tổng lập phương các nghiệm là \(2^3=8\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

\(\sqrt{9x^2+33x+28}+5\sqrt{4x-3}=5\sqrt{3x+4}+\sqrt{12x^2+19x-21}\)

\(\Leftrightarrow\sqrt{\left(3x+4\right)\left(3x+7\right)}+5\sqrt{4x-3}=5\sqrt{3x+4}+\sqrt{\left(3x+7\right)\left(4x-3\right)}\)

\(\Leftrightarrow\sqrt{\left(3x+4\right)\left(3x+7\right)}-5\sqrt{3x+4}=\sqrt{\left(3x+7\right)\left(4x-3\right)}-5\sqrt{4x-3}\)

\(\Leftrightarrow\sqrt{3x+4}\left(\sqrt{3x+7}-5\right)=\sqrt{4x-3}\left(\sqrt{3x+7}-5\right)\)

\(\Leftrightarrow\sqrt{3x+4}\left(\sqrt{3x+7}-5\right)-\sqrt{4x-3}\left(\sqrt{3x+7}-5\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x+7}-5\right)\left(\sqrt{3x+4}-\sqrt{4x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{3x+7}=5\\\sqrt{3x+4}=\sqrt{4x-3}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x+7=25\\3x+4=4x-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\) (thỏa mãn). Suy ra tổng các nghiệm của pt là \(6+7=13\)

Lập lên có

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

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

\(\Leftrightarrow-3\sqrt[3]{5}\sqrt[3]{2x+4}+3\sqrt[3]{5^2}\sqrt[3]{2x+4}+2x-1=2x\)

\(\Leftrightarrow3\sqrt[3]{5^2}\sqrt[3]{2x+4}-3\sqrt[3]{5}\left(2x+4\right)^{\frac{2}{3}}=0\)

\(\Leftrightarrow-3\sqrt[3]{5^2}\sqrt[3]{2x+4}\left(\sqrt[3]{2x+4}-\sqrt[3]{5}\right)=0\)

\(\Leftrightarrow x=\frac{1}{2}\left(tm\right)\)

chuyển vế lập phương là xong

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

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

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)