(x-2)^2=1
Tìm x
bài 1tìm x
X x 2 + X x 3 = 100
\(\left(x-2\right)\left(x^2+2x+4\right)+3x-4=\left(x+2\right)\left(x^2-2x+4\right)-x+1\)
\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)
\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)
\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)
\(\Rightarrow4x=21\)
\(\Rightarrow x=\dfrac{21}{5}\)
x^3+3x^2+3x=-1
Tìm x
\(\Rightarrow x^3+3x^2+3x+1=0\\ \Rightarrow\left(x+1\right)^3=0\Rightarrow x+1=0\Rightarrow x=-1\)
Cho x > y > 0; xy = 1
Tìm GTNN của A = \(\dfrac{x^2+y^2}{x-y}\).
Ta có x2+y2 / x-y = x2-2xy+y2+2xy / x-y
= (x-y)2+2xy / x-y
Mà xy = 1 => 2xy = 2. Thay vào, ta có
(x-y)2+2xy / x-y = (x-y)2+2 / x-y = (x-y)2 / x-y + 2 / x-y
= x-y + 2 / x-y
Áp dụng BĐT Cauchy, ta có
x-y + 2 / x-y ≥ 2.√(x-y).2 / x-y] = 2.√2 = (√2)3
Vậy Min A = (√2)3
(x+1/2)^2=1
tìm x ạ giúp minh với mình cần gấp ạ
\(\left(x+\dfrac{1}{2}\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=1\\x+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
tìm gtln của -3x^2+5x+6; -4x^2+4x-1
tìm gtnn của x^2+4x+7;x^2-x+1
Bài 2:
a: Ta có: \(x^2+4x+7\)
\(=x^2+4x+4+3\)
\(=\left(x+2\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=-2
Cho (x+\(\sqrt{y^2+1}\))(y+\(\sqrt{x^2+1}\))=1
Tìm GTNN của P=2(x2+y2)+x+y
Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\left(\dfrac{b^2-1}{2b}\right)+1}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\left(\dfrac{a^2-1}{2a}\right)+1}\right)=1\)
\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)
\(\Rightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4\left(ab\right)^2}+\dfrac{\left(a-b\right)^2}{4ab}=0\)
\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)
\(\Rightarrow\left(1-\dfrac{1}{ab}\right)\left(\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right)=0\)
\(\Rightarrow1-\dfrac{1}{ab}=0\Rightarrow ab=1\)
\(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Rightarrow x+y=0\Rightarrow y=-x\)
\(P=2\left(x^2+\left(-x\right)^2\right)+0=4x^2\ge0\)
Dấu "=" xảy ra khi \(x=y=0\)
1tìm x,y
(x2+y2)+(1-x)(1+y)
(x - 1) /9 + 1/3 = 1/y+2 và x -y = 1
Tìm x và y
Ta có x - y = 1 => x = y + 1
\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\Rightarrow\left(x+2\right)\left(y+2\right)=9\)
\(\Leftrightarrow\left(3+y\right)\left(y+2\right)=9\Leftrightarrow y^2+5y-3=0\Leftrightarrow y=\dfrac{-5\pm\sqrt{37}}{2}\)
thay vào tìm x
ps nhưng số xấu quá bạn ạ, kiểm tra lại đề nhé
ĐKXĐ:\(y\ne-2\)
\(\left\{{}\begin{matrix}\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+1-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y+3\right)\left(y+2\right)=9\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y+6-9=0\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y-3=0\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=\dfrac{-5+\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\\x=y+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{37}}{2}\\y=\dfrac{-5+\sqrt{37}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-3-\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\end{matrix}\right.\)
Ta có:
\(x-y=1\Rightarrow x=1+y\)
Thay vào
\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y}+2\) \(\left(đk:y\ne0\right)\)
\(\dfrac{x+2}{9}=\dfrac{2y+1}{y}\)
\(\Leftrightarrow\dfrac{y+3}{9}=\dfrac{2y+1}{y}\)
\(\Leftrightarrow y^2+3y=18y+9\)
\(\Leftrightarrow y^2-15y-9=0\)
\(\Leftrightarrow\)\(\left(y-\dfrac{15}{2}\right)^2=\dfrac{261}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}y-\dfrac{15}{2}=\dfrac{\sqrt{261}}{2}\\y-\dfrac{15}{2}=-\dfrac{\sqrt{261}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{\sqrt{261}+15}{2}\\y=\dfrac{15-\sqrt{261}}{2}\end{matrix}\right.\)
A =\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\) với ≥0,x≠1
tìm GTNN của A
$\large A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}=2-\frac{3}{\sqrt{x}+1}$
Ta có: $\large \sqrt{x}+1\ge1\Leftrightarrow -\frac{3}{\sqrt{x}+1}\ge-3$
Do đó: $\large A \ge 2-3=-1$
Vậy $A_{min}=-1$
Dấu $"="$ xảy ra khi $x=0$