\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)

<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1

<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)

<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0

<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0

<=> x + 66 = 0

<=> x = -66

hình như bn sai đề đúng ko

Đặt \(\hept{\begin{cases}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{cases}}\)

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

\(\Leftrightarrow\left(b-a\right)\left(4b-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=4b\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt[3]{65+x}=\sqrt[3]{65-x}\\\sqrt[3]{65+x}=4\sqrt[3]{65-x}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}65+x=65-x\\65+x=4\left(65-x\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=39\end{cases}}\)

\(xxx+xx+x+x=992\)

\(111x+11x+x+x=992\)

\(x\times\left(111+11+1+1\right)=992\)

\(x\times124=992\)

\(x=992\div124\)

\(x=8\)

\(4725+xxx+xx+x=54909\)

\(4725+111x+11x+x=54909\)

\(4725+x\times\left(111+11+1\right)=54909\)

\(4725+x\times123=54909\)

\(x\times123=54909-4725\)

\(x\times123=20184\)

Tự tính x

xx+x+5 =65 nha

b, ĐKXĐ: \(x\ge\frac{5}{2}\)

\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

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

\(\Leftrightarrow\sqrt{2x-5}=3\)

\(\Leftrightarrow x=7\left(tm\right)\)

a, ĐKXĐ: \(x\ge5\)

\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)

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

\(\Leftrightarrow2\sqrt{x-5}+6=0\)

\(\Leftrightarrow\sqrt{x-5}=-3\)

Phương trình vô nghiệm

a) Đặt \(\sqrt[3]{65+x}=a;\sqrt[3]{65-x}=b\)

Nhận xét x = 65 không phải là nghiệm. Xét x khác 65 thì \(b\ne0\)

PT \(\Leftrightarrow a^2+b^2-5ab=0\)

\(\Leftrightarrow\left(\frac{a}{b}\right)^2-5\left(\frac{a}{b}\right)+1=0\Leftrightarrow t^2-5t+1=0\left(\text{đặt }t=\frac{a}{b}\right)\)

Hình như chị ghi đề sai, số quá xấu:((

a/ Nghiệm xấu quá

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\) ta được:

\(a^2+b^2=5ab\Leftrightarrow a^2-5ab+b^2=0\)

\(\Leftrightarrow\left(a-\frac{5+\sqrt{21}}{2}b\right)\left(a-\frac{5-\sqrt{21}}{2}b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{5+\sqrt{21}}{2}b\\a=\frac{5-\sqrt{21}}{2}b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{65+x}=\frac{5+\sqrt{21}}{2}\sqrt[3]{65-x}\\\sqrt[3]{65+x}=\frac{5-\sqrt{21}}{2}\sqrt[3]{65-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}65+x=\left(\frac{5+\sqrt{21}}{2}\right)^3\left(65-x\right)\\65+x=\left(\frac{5-\sqrt{21}}{2}\right)^3\left(65-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(56+12\sqrt{21}\right)x=65\left(54+12\sqrt{21}\right)\\\left(56-12\sqrt{21}\right)x=65\left(54-12\sqrt{21}\right)\end{matrix}\right.\) \(\Rightarrow x=...\)

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

\(\Leftrightarrow3x-6+3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)=3x-6-6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=8\Rightarrow x=2\\\left(x-5\right)\left(2x-1\right)=-8\left(3x+2\right)\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x^2-13x+21=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{7}{2}\end{matrix}\right.\)

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Lập phương 2 vế ta đc

\(\left(65+x\right)^2+64\left(65-x\right)^2+3\sqrt[3]{64\left(65-x\right)^2\left(65+x\right)^x}.\left(\sqrt[3]{\left(65+x\right)^2}+\sqrt[3]{\left(65-x\right)^2}\right)=125\left(65^2-x^2\right)\)

<=>\(65x^2-8190x+274625+3\sqrt[3]{64\left(65^2-x^2\right)}.\sqrt[3]{65^2-x^2}=125\left(65^2-x^2\right)\)\(65x^2-8190x+274625+3.4.\sqrt[3]{65^2-x^2}=125\left(65^2-x^2\right)\)

Đặt 

 \(\sqrt[3]{\left(65+x\right)}=a;\sqrt[3]{65-x}=b\) => \(a^3+b^3=130\)  ta có Hpt :

\(a^2+4b^2=5ab\) (1) 

\(a^3+b^3=130\) (2)

từ pt (1) => a = b Hoặc a = 4b 

Thay vào pt (2) tìm ra b => a