ĐKXĐ: \(x\ge-2\)

Pt cho \(\Leftrightarrow4\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=x^2+x+10\)

Đặt \(\sqrt{x+2}=a;\sqrt{x^2-2x+4}=b\left(a,b\ge0\right)\)

Khi đó ta được pt: \(4ab=b^2+3a^2\Leftrightarrow\left(b-a\right)\left(b-3a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\b=3a\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x+2}=\sqrt{x^2-2x+4}\left(1\right)\\\sqrt{x^2-2x+4}=3\sqrt{x+2}\left(2\right)\end{cases}}\)

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

\(\left(2\right)\Leftrightarrow x^2-11x-14=0\Leftrightarrow\left(x-\frac{11}{2}\right)^2=\frac{177}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11+\sqrt{177}}{2}\\x=\frac{11-\sqrt{177}}{2}\end{cases}\left(tm\right)}\)

Vậy tập nghiệm của pt là \(S=\left\{1;2;\frac{11\pm\sqrt{177}}{2}\right\}.\)

Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

\(ĐKXĐ:x\ne5,8\)

\(\frac{6}{x-5}+\frac{x+2}{x-8}=\frac{18}{\left(x-5\right)\left(8-x\right)}-1\)

\(\Rightarrow\frac{6}{x-5}+\frac{x+2}{x-8}=-\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)

\(\Rightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)

\(\Rightarrow x^2+3x-58=-x^2+13x-58\)

\(\Rightarrow2x^2-10x=0\)

\(\Rightarrow2x\left(x-5\right)=0\)

\(\Rightarrow x\in\left\{0,5\right\}\)

Lời giải:
ĐKXĐ: $-10\leq x\leq 8$

$x^2+2x+7=(x+1)^2+6\geq 6(1)$

Áp dụng BĐT Bunhiacopxky:

$(\sqrt{8-x}+\sqrt{x+10})^2\leq (8-x+x+10)(1+1)=36$

$\Rightarrow \sqrt{8-x}+\sqrt{x+10}\leq 6(2)$

Từ $(1); (2)\Rightarrow \sqrt{8-x}+\sqrt{x+10}\leq 6\leq x^2+2x+7$

Để pt xảy ra thì $\sqrt{8-x}+\sqrt{x+10}=6=x^2+2x+7$

$\Leftrightarrow x=-1$

ĐKXĐ : -10 \(\le x\le8\)

Ta có \(3\sqrt{8-x}+3\sqrt{10+x}\le\dfrac{3^2+8-x}{2}+\dfrac{3^2+10+x}{2}=18\)

 (BĐT Cauchy)

=> \(\sqrt{8-x}+\sqrt{10+x}\le6\)

=> VT \(\le6\) (1)

Lại có VP = x² + 2x + 7 = (x + 1)² + 6 \(\ge6\) (2)

Từ (1) (2) => Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3=\sqrt{8-x}\\3=\sqrt{10+x}\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\)

Vậy x = -1 là nghiệm phương trình 

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\) = 5

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)

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

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

Nếu \(\sqrt{x-1}\ge2\Rightarrow\left|\sqrt{x-1}-2\right|=\sqrt{x-1}-2\Rightarrow\sqrt{x-1}-2+\sqrt{x-1}+3=5\)

\(\Rightarrow2\sqrt{x-1}=4\Leftrightarrow x=5\)

Nếu \(0\le\sqrt{x-1}< 2\Rightarrow\left|\sqrt{x-1}-2\right|=2-\sqrt{x-1}\Rightarrow2-\sqrt{x-1}+\sqrt{x-1}+3=5\)

\(\Leftrightarrow2+3=5\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)