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Ta có: \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}\)
Lại có: \(4\sqrt{x}\ge0\) với mọi x
\(3\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]>0\) với mọi x
\(\Rightarrow\) \(\dfrac{4\sqrt{x}}{3\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}\ge0\) với mọi x
Dấu "=" xảy ra \(\Leftrightarrow\) x = 0
Vậy ...
Chúc bn học tốt! (Mk ms nghĩ ra được GTNN thôi thông cảm!)
Còn tìm GTLN:
Ta có: \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left[\left(\sqrt{x}-1\right)^2+\sqrt{x}\right]}\le\dfrac{4\sqrt{x}}{3\sqrt{x}}=\dfrac{4}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\) \(\sqrt{x}-1=0\) \(\Leftrightarrow\) x = 1
Vậy ...
Chúc bn học tốt!
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+2\sqrt{2\left(x-2\right)}}+\sqrt{x-2\sqrt{2\left(x-2\right)}}=2\sqrt{2}\)
\(\Leftrightarrow2x+2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8\)
\(\Leftrightarrow2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8-2x\)
\(\Leftrightarrow4\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]=64-32x+4x^2\)
\(\Leftrightarrow4x^2-32x+64=64-32x+4x^2+\)
\(\Leftrightarrow64=64\) (Đúng)
⇒ Phương trình có vô số nghiệm.
Vậy \(S=\mathbb R\).
Giải phương trình sau:
\(\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}=4}\)
Các cao nhân hỗ trợ em với ạ~
ĐK:....
\(\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4\)
\(\Leftrightarrow\left(\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}\right)\left(\sqrt{x+\sqrt{x+11}}-\sqrt{x-\sqrt{x+11}}\right)=4\left(\sqrt{x+\sqrt{x+11}}-\sqrt{x-\sqrt{x+11}}\right)\)
\(\Leftrightarrow x+\sqrt{x+11}-x+\sqrt{x+11}=4\left(\sqrt{x+\sqrt{x+11}}-\sqrt{x-\sqrt{x+11}}\right)\)
\(\Leftrightarrow2\sqrt{x+11}=4\sqrt{x+\sqrt{x+11}}-4\sqrt{x-\sqrt{x+11}}\)
\(\Leftrightarrow2\left(\sqrt{x+\sqrt{x+11}}-\sqrt{x-\sqrt{x+11}}\right)=\sqrt{x+11}\)
\(\Leftrightarrow4\left(x+\sqrt{x+11}+x-\sqrt{x+11}-2\sqrt{\left(x+\sqrt{x+11}\right)\left(x-\sqrt{x+11}\right)}\right)=x+11\)
\(\Leftrightarrow4\left(2x-2\sqrt{x^2-x-11}\right)=x+11\)
\(\Leftrightarrow8x-8\sqrt{x^2-x-11}=x+11\)
\(\Leftrightarrow8\sqrt{x^2-x-11}=7x-11\)
\(\Leftrightarrow64\left(x^2-x-11\right)=\left(7x-11\right)^2\)
\(\Leftrightarrow64x^2-64x-704=49x^2-154x+121\)
\(\Leftrightarrow15x^2+90x-825=0\)
\(\Leftrightarrow15x^2-75x+165x-825=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(chon\right)\\x=-11\left(loai\right)\end{matrix}\right.\)
Vậy...
1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(B=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x+2}{\sqrt{x}}\)
Nếu không phiền, bạn có thể giải chi tiết cho mình được không ạ. Mình cảm ơn nhiều !
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
Lời giải:
ĐK:.............
Đặt $\sqrt{2x^2+x+6}=a; \sqrt{x^2+x+2}=b$ với $a,b\geq 0$ thì PT trở thành:
$a+b=\frac{a^2-b^2}{x}$
$\Leftrightarrow (a+b)(\frac{a-b}{x}-1)=0$
Nếu $a+b=0$ thì do $a,b\geq 0$ nên $a=b=0$
$\Leftrightarrow \sqrt{2x^2+x+6}=\sqrt{x^2+x+2}=0$ (vô lý)
Nếu $\frac{a-b}{x}-1=0$
$\Leftrightarrow a-b=x$
$\Leftrightarrow \sqrt{2x^2+x+6}=\sqrt{x^2+x+2}+x$
$\Rightarrow 2x^2+x+6=2x^2+x+2+2x\sqrt{x^2+x+2}$ (bình phương 2 vế)
$\Leftrightarrow 2=x\sqrt{x^2+x+2}(1)$
$\Rightarrow 4=x^2(x^2+x+2)$
$\Leftrightarrow x^4+x^3+2x^2-4=0$
$\Leftrightarrow (x-1)(x^3+2x^2+4x+4)=0$
Từ $(1)$ ta có $x>0$. Do đó $x^3+2x^2+4x+4>0$ nên $x-1=0$
$\Rightarrow x=1$Vậy..........