Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) ĐKXĐ: \(x\ne1\)
Ta có:
\(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}-2=0\)
\(\Leftrightarrow\left(x+\frac{x}{x-1}\right)^3-3x.\frac{x}{x-1}\left(x+\frac{x}{x-1}\right)+\frac{3x^2}{x-1}-2=0\)
\(\Leftrightarrow\left(\frac{x^2}{x-1}\right)^3-3\left(\frac{x^2}{x-1}\right)^2+\frac{3x^2}{x-1}-2=0\)
Đặt \(\frac{x^2}{x-1}=a\)
Khi đó pt đã cho trở thành:
\(a^3-3a^2+3a-2=0\)
\(\Leftrightarrow\left(a-1\right)^3=1\Rightarrow a-1=1\Leftrightarrow a=2\)
Theo cách đặt: \(\frac{x^2}{x-1}=2\Rightarrow x^2=2x-2\Leftrightarrow x^2-2x+1=-1\Leftrightarrow\left(x-1\right)^2=-1\left(ptvn\right)\)
a) ĐKXĐ: \(x\ge8\)
Ta có:
\(x-\sqrt{x-8}-3\sqrt{x}+1=0\)
\(\Leftrightarrow x-9-\left(\sqrt{x-8}-1\right)-3\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x-9-\frac{x-9}{\sqrt{x-8}+1}-3.\frac{x-9}{\sqrt{x}+3}=0\)
\(\Leftrightarrow\left(x-9\right)\left(\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1=0\end{cases}}\)
+) \(x-9=0\Leftrightarrow x=9\left(TMĐKXĐ\right)\)
+) \(\frac{3}{\sqrt{x}+3}=\frac{\sqrt{x-8}}{\sqrt{x-8}+1}\Rightarrow\sqrt{x\left(x-8\right)}=3\)
\(\Leftrightarrow x^2-8x-9=0\Leftrightarrow\orbr{\begin{cases}x=9TMĐKXĐ\\x=-1\left(KTMĐKXĐ\right)\end{cases}}\)
Vaayh pt có 1 nghiệm là x=9
b, Ta có
\(\frac{\sqrt{x}+1}{y+1}=\frac{\left(\sqrt{x}+1\right)\left(y+1\right)-y-y\sqrt{x}}{y+1}=\sqrt{x}+1-\frac{y\left(\sqrt{x}+1\right)}{y+1}\)
Mà \(y+1\ge2\sqrt{y}\)
=> \(\frac{\sqrt{x}+1}{y+1}\ge\sqrt{x}+1-\frac{1}{2}\sqrt{y}\left(\sqrt{x}+1\right)\)
Khi đó
\(P\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3-\frac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=3\)
=> \(P\ge\frac{1}{2}.3+3-\frac{3}{2}=3\)
Vậy MinP=3 khi x=y=z=1
Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ
Mn giúp em vs ạ! Thanks trước!
ráng làm nốt rồi đi ngủ thoyy
1.
a) ĐK: \(x\ge2\)
\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)
Vậy...
b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)
\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)
Vậy...
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
Nhân cả 2 vế với \(\sqrt{2}\) ta được :
\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)
2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)
Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)
Tương tự 2 trường hợp còn lại ta đều được \(B=0\)
Vậy \(B=0\)
ĐKXĐ: z>0
pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)
<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)
<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)
<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)
<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)
<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)
<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)
vậy x=2
