CÁi  này easy mà .-.

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

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

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

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

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

Nguyễn Việt Lâm, Nguyễn Lê Phước Thịnh giúp vs!

ĐKXĐ: \(x\ge2\).

Với \(x\ge2\) ta có \(VP\le2;VT\ge2\).

Do đó nghiệm của pt là \(x=2\).

mk đánh đề bị lộn nha

pt đó chỉ bằng 2x thuj

â) \(\sqrt{x+9}=7\\ \Rightarrow x+9=49\\ \Rightarrow x=40\)

b) \(\sqrt{x-4}=4-x\\ \Rightarrow x-4=16-8x+x^2\\ \Rightarrow x^2-9x+20=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

c) \(\sqrt{x^2-12x+36}=81\\ \Rightarrow x-6=81\\ \Rightarrow x=87\)

a: Ta có: \(\sqrt{x+9}=7\)

\(\Leftrightarrow x+9=49\)

hay x=40

b: Ta có: \(\sqrt{x-4}=4-x\)

\(\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-12x+36}=81\)

\(\Leftrightarrow\left|x-6\right|=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

Sửa đề: \(19 + 3x + 4\sqrt{-x^2 - x + 6} = 10\sqrt{2 - x} + 10\sqrt{x + 3}\) (1)

ĐKXĐ: -3<=x<=2

Đặt \(u=\sqrt{2-x};v=\sqrt{x+3}\) (Điều kiện: u>0; v>0)

\(u^2+v^2=2-x+3+x=5\)

\(uv=\sqrt{\left(2-x\right)\left(x+3\right)}=\sqrt{2x+6-x^2-3x}=\sqrt{-x^2-x+6}\)

\(v^2=x+3\)

=>\(x=v^2-3\)

=>\(3x=3v^2-9\)

(1) sẽ trở thành: \(19 + (3v^2 - 9) + 4uv = 10u + 10v\)

=>\(10+3v^2+4uv-10u-10v=0\)

=>\(u(4v-10)=-3v^2+10v-10=-3\left(v^2-\frac{10}{3}v+\frac{10}{3}\right)=-3\left(v^2-2\cdot v\cdot\frac53+\frac{25}{9}+\frac59\right)=-3\left(v-\frac53\right)^2-\frac53<0\) ∀v

=>u(4v-10)<0

=>4v-10<0

=>4v<10

=>v<5/2

=>\(u = \frac{3v^2 - 10v + 10}{10 - 4v}\)

=>\(5 - v^2 = \left( \frac{3v^2 - 10v + 10}{10 - 4v} \right)^2\)

=>\((5 - v^2)(10 - 4v)^2 = (3v^2 - 10v + 10)^2\)

=>\((5 - v^2)(16v^2 - 80v + 100) = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(-16v^4 + 80v^3 - 20v^2 - 400v + 500 = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(25v^4 - 140v^3 + 180v^2 + 200v - 400 = 0\)

=>\(5v^4 - 28v^3 + 36v^2 + 40v - 80 = 0\)

=>\((v - 2)(5v^3 - 18v^2 + 40) = 0\)

=>v-2=0

=>v=2

=>x+3=4

=>x=1(nhận)

ĐKXĐ: 5-x>=0 và x+8>=0

=>-8<=x<=5

Ta có: \(13\sqrt{5-x}+18\sqrt{x+8}=61+x+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\sqrt{5-x}-26+18\sqrt{x+8}-54=x-19+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\left(\sqrt{5-x}-2\right)+18\left(\sqrt{x+8}-3\right)=x-1-18+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\frac{5-x-4}{\sqrt{5-x}+2}+18\cdot\frac{x+8-9}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{\left(5-x\right)\left(x+8\right)}-6\right)\)

=>\(13\cdot\frac{1-x}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{5x+40-x^2-8x}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{-x^2-3x+40}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\cdot\frac{-x^2-3x+40-36}{\sqrt{-x^2-3x+40}+6}\)

=>(x-1)\(\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-3x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-4x+x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{\left(x+4\right)\left(-x+1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1-3\cdot\frac{\left(x+4\right)\left(x-1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}-1+3\cdot\frac{x+4}{\sqrt{-x^2-3x+40}+6}\right)=0\)

=>x-1=0

=>x=1(nhận)

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

+) Lập phương 2 vế ta được :

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

Thay ( 1 ) vào ( 2 ) ta có : 

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

\(\Rightarrow4x^3-5x=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\pm\frac{\sqrt{5}}{2}\end{cases}}\)

P/s : ko có tgian làm full . Thông cảm nhen ^-^