Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Hoàng Như Quỳnh]

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

\(ĐKXĐ:x\ge1\)

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

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

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

\(\orbr{\begin{cases}2-\sqrt{x-1}+\sqrt{x-1}+3=5\\\sqrt{x-1}-2+\sqrt{x-1}+3=5\end{cases}\orbr{\begin{cases}5=5\left(TM\forall x\right)\\\sqrt{x-1}=2\end{cases}\orbr{\begin{cases}5=5\\x=5\left(TM\right)\end{cases}}}}\)

vậy pt có nghiệm là \(1\le x\le+\infty\)

\(2,\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)

\(ĐKXĐ:x\ge\frac{3}{2}\)

\(\sqrt{x+\sqrt{6}\sqrt{x-\frac{3}{2}}}+\sqrt{x-\sqrt{6}\sqrt{x-\frac{3}{2}}}=\sqrt{6}\)

\(\sqrt{x-\frac{3}{2}+\sqrt{6}\sqrt{x-\frac{3}{2}}+\frac{3}{2}}+\sqrt{x-\frac{3}{2}-\sqrt{6}\sqrt{x-\frac{3}{2}}+\frac{3}{2}}=\sqrt{6}\)

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

\(\left|\sqrt{x-\frac{3}{2}}+\frac{3}{2}\right|+\left|\sqrt{x-\frac{3}{2}}-\frac{3}{2}\right|=\sqrt{6}\)

\(\orbr{\begin{cases}\sqrt{x-\frac{3}{2}}+\frac{3}{2}+\frac{3}{2}-\sqrt{x-\frac{3}{2}}=\sqrt{6}\\\sqrt{x-\frac{3}{2}}+\frac{3}{2}-\frac{3}{2}+\sqrt{x-\frac{3}{2}}=\sqrt{6}\end{cases}}\orbr{\begin{cases}3=\sqrt{6}\\2\sqrt{x-\frac{3}{2}}=\sqrt{6}\end{cases}}\)

\(\orbr{\begin{cases}3=\sqrt{6}\left(KTM\right)\\x=3\left(TM\right)\end{cases}}\)

\(3,\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

\(ĐKXĐ:x\ge3\)

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

\(\frac{x^2-5x+4}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}=\frac{x-4}{\sqrt{x-2}+\sqrt{2}}+\frac{x^2-2x-8}{\sqrt{x^2-2x-3}+\sqrt{5}}\)

\(\frac{\left(x-4\right)\left(x-1\right)}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{x-4}{\sqrt{x+1}+\sqrt{5}}=\frac{x-4}{\sqrt{x-2}+\sqrt{2}}+\frac{\left(x-4\right)\left(x+2\right)}{\sqrt{x^2-2x-3}+\sqrt{5}}\)

\(\left(x-4\right)\left(\frac{x-1}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{1}{\sqrt{x+1}+\sqrt{5}}-\frac{1}{\sqrt{x-2}+\sqrt{2}}-\frac{x+2}{\sqrt{x^2-2x-3}+\sqrt{5}}\right)=0\)

\(\orbr{\begin{cases}x=4\left(TM\right)\\\frac{x-1}{\sqrt{x^2-5x+6}+\sqrt{2}}+\frac{1}{\sqrt{x+1}+\sqrt{5}}-\frac{1}{\sqrt{x-2}+\sqrt{2}}-\frac{x+2}{\sqrt{x^2-2x-3}-\sqrt{5}}=0\end{cases}}\)

bạn lập luận cái dưới vô nghiệm 

Answer 2

[Victorique de Blois]

10, \(đk:x\ge\frac{1}{2}\)

\(\sqrt{x+3}+\sqrt{2x-1}=3\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{2x-1}-3=0\)

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

\(\Leftrightarrow\frac{x+3-4}{\sqrt{x+3}+2}+\frac{2x-1-1}{\sqrt{2x-1}+1}=0\)

\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{2x-2}{\sqrt{2x-1}+1}=0\)

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

với x >= 1/2 thì ngoặc thứ 2 > 0

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

8, đk \(\orbr{\begin{cases}x\ge0\\x\le-8\end{cases}}\)

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

\(\Leftrightarrow x\left(x+8\right)-3=2\sqrt{x\left(x+8\right)}\)

đặt \(\sqrt{x\left(x+8\right)}=a\left(a\ge0\right)\)

pt trở thành : \(a^2-3=2a\Leftrightarrow a^2-2a-3=0\)

\(\Leftrightarrow\left(a-3\right)\left(a+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-1\left(loai\right)\end{cases}}\)

a = 3 => \(\sqrt{x\left(x+8\right)}=3\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-9\left(tm\right)\end{cases}}\)

7, đk \(x>0\)

\(\sqrt{\frac{x^2+x+1}{x}}+\sqrt{\frac{x}{x^2+x+1}}=\frac{7}{4}\)

\(\Leftrightarrow\frac{x^2+x+1}{x}+\frac{x}{x^2+x+1}+2\sqrt{\frac{x^2+x+1}{x}\cdot\frac{x}{x^2+x+1}}=\frac{49}{16}\)

\(\Leftrightarrow\frac{x^4+x^2+1+2x^3+2x^2+2x+x^2}{x\left(x^2+x+1\right)}+2=\frac{49}{16}\)

\(\Leftrightarrow\frac{x^4+2x^3+4x^2+2x+1}{x\left(x^2+x+1\right)}=\frac{17}{16}\)

\(\Leftrightarrow16x^4+32x^3+64x^2+32x+16=17x^3+17x^2+17x\)

\(\Leftrightarrow16x^4+15x^3+47x^2+15x+16=0\)

bấm mt nó ra nghiệm ảo :v