Bài 9 giải phương trình
a) \(\sqrt{4}x-20+\sqrt{x}-5-\frac{1}{3}\sqrt{9x-45}=4\)
b)\(\sqrt{x^2-5x+16}=x+2\)
c) \(\frac{\sqrt{x}+1}{=\sqrt{x}-1}=2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đk: \(x\ge0\)
Xét x = 0 thấy thỏa mãn
Xét x > 0 thì: \(x\sqrt{x}-2\sqrt{x}-x=0\)
\(\Leftrightarrow x-\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
Mà \(\sqrt{x}+2>0\left(\forall x\right)\Rightarrow\sqrt{x}-1=0\Rightarrow x=1\)
Vậy x = 1 hoặc x = 0
đk: \(x\ge0\)
Ta có:
\(A=x-4\sqrt{x}+10\)
\(A=\left(x-4\sqrt{x}+4\right)+6\)
\(A=\left(\sqrt{x}-2\right)^2+6\ge6\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(\sqrt{x}-2\right)^2=0\Rightarrow x=4\)
Vậy Min(A) = 6 khi x = 4
a) đk: \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
Ta có:
\(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=\frac{2}{x-1}\)
b) Để M nguyên => \(\left(x-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Leftrightarrow x\in\left\{-1;0;2;3\right\}\) mà theo đk nên \(x\in\left\{2;3\right\}\)
AD=\(\sqrt{3}\)R
AE=(1/4)AD=\(\frac{\sqrt{3}}{4}\)R
DE=\(\frac{3\sqrt{3}}{4}\)R
Cho số thực x thỏa mãn \(0\le x\le5\). Tìm GTLN,GTNN của
\(P=x\sqrt{8-x}+\left(5-x\right)\sqrt{x+3}\)
\(P^2=\left(x\sqrt{8-x}+\left(5-x\right)\sqrt{x+3}\right)^2\)
\(=x^2\left(8-x\right)+\left(5-x\right)^2\left(x+3\right)+2x\left(5-x\right)\sqrt{8-x}\sqrt{x+3}\)
\(=x^2-5x+75+2x\left(5-x\right)\sqrt{8-x}\sqrt{x+3}\)
Có \(\sqrt{8-x}\sqrt{x+3}\le\frac{8-x+x+3}{2}=\frac{11}{2}\)
\(0\le x\le5\Rightarrow x\left(5-x\right)\ge0\)
Suy ra \(P^2\le x^2-5x+75+2x\left(5-x\right).\frac{11}{2}\)
\(=x^2-5x+75+11x\left(5-x\right)\)
\(=10x\left(5-x\right)+75\)
\(\le10.\left(\frac{x+5-x}{2}\right)^2+75=\frac{275}{2}\)
Suy ra \(P\le\sqrt{\frac{275}{2}}=\frac{5\sqrt{22}}{2}\)
Dấu \(=\)xảy ra khi \(\hept{\begin{cases}8-x=x+3\\x=5-x\end{cases}}\Leftrightarrow x=\frac{5}{2}\).
Vậy \(maxP=\frac{5\sqrt{22}}{2}\).
\(P^2=x\left(x-5\right)+75+2x\left(5-x\right)\sqrt{8-x}\sqrt{x+3}\)
\(=x\left(5-x\right)\left(2\sqrt{8-x}\sqrt{x+3}-1\right)+75\)
Có \(0\le x\le5\)nên \(\sqrt{8-x}\ge\sqrt{8-5}>1,\sqrt{x+3}\ge\sqrt{0+3}>1\)
suy ra \(\sqrt{8-x}\sqrt{x+3}>1\Rightarrow2\sqrt{8-x}\sqrt{x+3}-1>0\)
\(0\le x\le5\) nên \(x\left(5-x\right)\ge0\)
Suy ra \(P^2=x\left(5-x\right)\left(2\sqrt{8-x}\sqrt{x+3}-1\right)+75\ge75\)
\(P\ge\sqrt{75}=5\sqrt{3}\).
Dấu \(=\)xảy ra khi \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\).
Vậy \(minP=5\sqrt{3}\).