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

\(\Leftrightarrow x-5+2x-1+3\sqrt[3]{x-5}\sqrt[3]{2x-1}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)=3x+2-8-3.2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)

\(\Leftrightarrow\sqrt[3]{2x^2-11x+5}\left(\sqrt[3]{x-5}+\sqrt[3]{2x-1}\right)=-2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\text{ (*)}\)

Mà: \(\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\)

\(+\text{Nếu }\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2=0\Leftrightarrow x=2\) thì pt thành 0 = 0 (thỏa mãn)

\(+\text{Xét }\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\ne0\)

\(\text{(*)}\Leftrightarrow\sqrt[3]{2x^2-11x+5}=-2\sqrt[3]{3x+2}\)

\(\Leftrightarrow2x^2-11x+5=-8\left(3x+2\right)\)

\(\Leftrightarrow2x^2+13x+21=0\)

\(\Leftrightarrow x=-3\text{ hoặc }x=-\frac{7}{2}\)

Vậy, tập nghiệm của phương trình là: \(S=\left\{-\frac{7}{2};\text{ }-3;\text{ }2\right\}\)

a. \(9x=225\Rightarrow x=25\)

b. \(2x=8\Rightarrow x=4\)

c. \(3\left(2x-3\right)=6\Rightarrow6x=15\Rightarrow x=\frac{15}{6}\)

d. \(4\left(x+1\right)=8\Rightarrow4x=4\Rightarrow x=1\)

e. \(\sqrt{x+2}.\sqrt{x-2}-\sqrt{x-2}=0\Rightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

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

hoặc \(\sqrt{x+2}-1=0\Rightarrow\sqrt{x+2}=1\Rightarrow x+2=1\Rightarrow x=-1\)

f. \(\sqrt{x+1}+3\sqrt{x+1}=4\Rightarrow4\sqrt{x+1}=4\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)

g. \(\sqrt{x-2}\left(1-\sqrt{x}\right)=0\)

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

hoặc \(1-\sqrt{x}=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\)

h. \(\sqrt{x+3}.\sqrt{x-3}-\sqrt{x+3}=0\Rightarrow\sqrt{x+3}\left(\sqrt{x-3}-1\right)=0\)

=> \(\sqrt{x+3}=0\Rightarrow x=-3\)

hoặc \(\sqrt{x-3}-1=0\Rightarrow\sqrt{x-3}=1\Rightarrow x=4\)

√x−2(1−√x)=0

=> √x−2=0⇒x−2=0⇒x=2

hoặc 1−√x=0⇒√x=1⇒x=1

h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0

=> √x+3=0⇒x=−3

hoặc 

√x−2(1−√x)=0

=> √x−2=0⇒x−2=0⇒x=2

hoặc 1−√x=0⇒√x=1⇒x=1

h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0

=> √x+3=0⇒x=−3

hoặc 

http://diendan.hocmai.vn/showthread.php?t=287459

a+b=2

muốn giải nhắn mình BnoHi facebook

thong cam bai nay tui dc lam bao h

\(a^4+b^4+\left(a+b\right)^2=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2+2ab\right)^2\)

\(=\left(a^2+b^2\right)-2a^2b^2+\left(a^2+b^2\right)+4ab\left(a^2+b^2\right)+4a^2b^2\)

\(=2\left[\left(a^2+b^2\right)^2+2ab\left(a^2+b^2\right)+a^2b^2\right]\)

\(=2\left(a^2+b^2+ab\right)^2\)

Tương tự: \(x^4+y^4+\left(x+y\right)^4=2\left(x^2+y^2+xy\right)^2\)

Mà \(a^2+b^2+\left(a+b\right)^2=x^2+y^2+\left(x+y\right)^2\Rightarrow2\left(a^2+b^2+ab\right)=2\left(x^2+y^2+xy\right)\)

\(\Rightarrow2\left(a^2+b^2+ab\right)^2=2\left(x^2+y^2+xy\right)^2\)

hay \(a^4+b^4+\left(a+b\right)^4=x^4+y^4+\left(x+y\right)^4\)