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a: 3x-2=2x-3
=>x=-1
b: 2x+3=5x+9
=>-3x=6
=>x=-2
c: 5-2x=7
=>2x=-2
=>x=-2
d: 10x+3-5x=4x+12
=>5x+3=4x+12
=>x=9
e: 11x+42-2x=100-9x-22
=>9x+42=78-9x
=>18x=36
=>x=2
f: 2x-(3-5x)=4(x+3)
=>2x-3+5x=4x+12
=>7x-3=4x+12
=>3x=15
=>x=5
\(\frac{\left(x+6\right)\left(x+6\right)}{2}-\frac{4x}{3}=0\)
\(\Leftrightarrow\frac{\left(x+6\right)^2}{2}=\frac{4x}{3}\)
\(\Leftrightarrow\frac{3\left(x+6\right)^2}{6}=\frac{8x}{6}\)
\(\Leftrightarrow3\left(x+6\right)^2=8x\)
\(\Leftrightarrow3\left(x^2+12x+36\right)-8x=0\)
\(\Leftrightarrow3x^2+36x+108-8x=0\)
\(\Leftrightarrow3x^2+28x+108=0\)
=> pt vô ngiệp
\(\frac{\left(x+6\right)^2}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{x^2+12x+36}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{3x^2+36x+108}{6}-\frac{8x}{6}=0\)
\(\Rightarrow\frac{3x^2+28x+108}{6}=0\)
\(\Rightarrow3x^2+28x+108=0\)
Ta có: \(\Delta=28^2-4.3.108=-512< 0\)
Vậy pt vô nghiệm
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\\ =\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(\left(x^2\right)^2-2x^2+1\right)+4}\\ =\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\)
do: \(+\left(x+1\right)^2\ge0\Rightarrow3.\left(x+1\right)^2+9\ge9\Rightarrow\sqrt{3\left(x+1\right)^2+9}\ge\sqrt{9}=3\)(1)\(+\left(x^2-1\right)^2\ge0\Rightarrow5\left(x^2-1\right)^2+4\ge4\Rightarrow\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{4}=2\)(2)
từ (1) và(2)\(\Rightarrow\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\)
câu b bạn làm tương tự
Ta có : \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)
\(\Leftrightarrow x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x-1=0\)
hoặc \(x^2+x+6=0\)
\(\Leftrightarrow\) \(x=-2\)(tm)
hoặc \(x=1\)(tm)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)
\(x^2-4xy+5y^2+10x-22y+28\)
\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(=\left(2y-x-5\right)^2+\left(y-1\right)^2+2\)
Ta có :
\(\left(2y-x-5\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2y-x-5\right)^2+\left(y-1\right)^2+2\ge2\forall x\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=0\\\left(2y-x-5\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=0\\2y-x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-3\end{matrix}\right.\)
Vậy biểu thức đạt GTNN = 2 ⇔ \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(4x^2+8x=0\)
\(\Leftrightarrow4x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
a) \(4x^2+8x=0\)
\(\Rightarrow4x\left(x+8\right)=0\)
\(\Rightarrow4x=0\) hoặc \(x+8=0\)
\(TH1:4x=0\Rightarrow x=4:0\Rightarrow x=0\)
\(TH2:x+8=0\Rightarrow x=0-8\Rightarrow x=-8\)
Vậy nghiệm của đa thức \(4x^2+8x=0\) là: \(\left\{0;-8\right\}\)
\(A=-x^2+4x+7=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\)
Ta thấy : \(-\left(x-2\right)^2+11\le11\)\(\Leftrightarrow maxA=11\)khi \(x=2\)
\(B=-4x^2+4x-5=-\left(4x^2-4x+1\right)-4=-\left(2x-1\right)^2-4\)
Ta thấy : \(-\left(2x-1\right)^2-4\le-4\)\(\Leftrightarrow maxB=-4\)khi \(x=\frac{1}{2}\)
\(C=-x^2+x+5=-\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)+\frac{21}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{21}{4}\)
Ta thấy : \(-\left(x-\frac{1}{2}\right)^2+\frac{21}{4}\le\frac{21}{4}\)\(\Leftrightarrow maxC=\frac{21}{4}\)khi \(x=\frac{1}{2}\)
tk mk nka !!!
thanks