Question

[Nguyễn Duy Long]

Xin nhờ mọi người:

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

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

3,\(x\sqrt{y-1}+2y\sqrt{x-1}=\frac{3xy}{2}\)

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

5.\(\left(\sqrt{7+\sqrt{48}}\right)^x+\left(7-\sqrt{48}\right)^x=14\)

6.\(\sqrt{17-x^2}=\left(3-\sqrt{x}\right)^2\)

Mỗi người giúp một ít, tích tiểu thành đại nha

Answer 1

[Thắng Nguyễn]

Tiếp =))

c)Áp dụng BĐT AM-GM ta có:

\(x\sqrt{y-1}\le\frac{x\left(y-1+1\right)}{2}=\frac{xy}{2}\)

\(2y\sqrt{x-1}\le\frac{2y\left(x-1+1\right)}{2}=\frac{2xy}{2}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=x\sqrt{y-1}+2y\sqrt{x-1}\le\frac{3xy}{2}=VP\)

Nên xảy ra khi \(x=y\) thay vào giải ra có: x=y=2

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

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

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

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

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

pt trong ngoặc vn nên x=1

Tắm đã làm nốt cho :))

Answer 2

[Thắng Nguyễn]

Chả ai giúp t gank =)), mà lần sau đăng ít 1 thôi đăng lắm thế này nhìn nản cmn luôn ấy

a)\(\sqrt{x^2+x-5}+\sqrt{-x^2+x+3}=x^2-3x+4\)

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

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

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

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

Pt trong ngoặc <0 nên x=2 là nghiệm

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

Đk:\(x\ge-\frac{1}{2}\)

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

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

\(\Leftrightarrow\frac{x^2-3x}{2}-\frac{2x^3-5x^2-3x}{\sqrt{2x^3-x^2+x+1}+2x+1}=0\)

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

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

Pt trong ngoặc vô nghiệm nốt nên 

\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)