Đk: x>=0​

\(\Leftrightarrow\frac{x^2-1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+3\frac{x-1}{\sqrt{x}+1}+\frac{x-1}{\sqrt{x+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+\frac{3}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}\right)=0\)

Với đk x>=0 ta có\(\frac{x+1}{\sqrt[3]{\left(x^2+26\right)^2}+3\sqrt[3]{x^2+26}+9}+\frac{3}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}>0\)

pt <=> x-1=0<=>x=1 (tm)

Do có quá ít câu hỏi nên bạn nào trả lời được, mình sẽ xóa khỏi mục "Câu hỏi hay" nhé!

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-3=1\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

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

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

a: ĐKXĐ: 3<=x<=8

\(\sqrt{\left(x-3\right)\left(8-x\right)}+26>-x^2+11x\)

=>\(\sqrt{8x-x^2-24+3x}+26>-x^2+11x\)

=>\(\sqrt{-x^2+11x-24}>-x^2+11x-26\) (1)

Đặt \(a=\sqrt{-x^2+11x-24}\) (ĐIều kiện: a>=0)

=>\(a^2=-x^2+11x-24\)

=>\(a^2-2=-x^2+11x-24-2=-x^2+11x-26\)

(1) sẽ trở thành: \(a>a^2-2\)

=>\(a^2-2

=>\(a^2-a-2<0\)

=>(a-2)(a+1)<0

=>a-2<0

=>a<2

=>\(-x^2+11x-24<4\)

=>\(-x^2+11x-28<0\)

=>\(x^2-11x+28>0\)

=>(x-7)(x-4)>0

=>x>7 hoặc x<4

kết hợp ĐKXĐ, ta được: 3<=x<4 hoặc 7<x<=8

b: ĐKXĐ: x∈R

\(\left(x+1\right)\left(x+4\right)<5\cdot\sqrt{x^2+5x+28}\)

=>\(x^2+5x+4<5\cdot\sqrt{x^2+5x+28}\) (2)

Đặt \(a=\sqrt{x^2+5x+28}\) (Điều kiện: a>0)

=>\(a^2=x^2+5x+28\)

=>\(a^2-24=x^2+5x+4\)

(2) sẽ trở thành: \(a^2-24<5a\)

=>\(a^2-5a-24<0\)

=>(a-8)(a+3)<0

=>a-8<0

=>a<8

=>\(x^2+5x+28<64\)

=>\(x^2+5x-36<0\)

=>(x+9)(x-4)<0

=>-9<x<4

\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)

\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)

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

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

\(2\sqrt{x+3}=x+4\)

\(4\left(x+3\right)=x^2+8x+14\)

\(x^2+4x+2=0\)

\(\Delta=16-8=8\)

\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)

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