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\(C=\left(x^2+\dfrac{y^2}{4}+4-xy+4x-2y\right)+\dfrac{3}{4}\left(y^2-4y+4\right)+1011\)
\(=\left(x-\dfrac{y}{2}+2\right)^2+\dfrac{3}{4}\left(y-2\right)^2+1011\ge1011\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-1;2\right)\)
a) Ta có: \(B=x^2+4y^2+4x-4y\)
\(=\left(x^2+4x+4\right)+\left(4y^2-4y+1\right)-5\)
\(=\left(x+2\right)^2+\left(2y-1\right)^2-5\ge-5\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-2;\dfrac{1}{2}\right)\)
\(A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\)
\(A=\left|x-1\right|+\left|4-x\right|+\left|x-2\right|+\left|3-x\right|\)
+) Đặt \(B=\left|x-1\right|+\left|4-x\right|\ge\left|x-1+4-x\right|=3\)
Dấu '' = '' xảy ra \(\Leftrightarrow\left(x-1\right)\left(4-x\right)=0\)
\(\Leftrightarrow1\le x\le4\)
+) Đặt \(C=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)
Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow2\le x\le3\)
\(\Rightarrow A=\left|x-1\right|+\left|4-x\right|+\left|x-2\right|+\left|3-x\right|\ge4\)
Dấu '' = '' xảy ra
\(\Leftrightarrow\hept{\begin{cases}1\le x\le4\\2\le x\le3\end{cases}\Leftrightarrow2\le x\le3}\)
Vậy.................
A=3x2-x+4
\(=3\left(x^2-\frac{x}{3}+\frac{4}{3}\right)\)
\(=3\left(x-\frac{1}{6}\right)^2+\frac{47}{12}\ge0+\frac{47}{12}=\frac{47}{12}\)
Dấu = khi \(x=\frac{1}{6}\)
Vậy MinA=\(\frac{47}{12}\Leftrightarrow x=\frac{1}{6}\)
B=(x-2)(x-5)(x2-7x-10)
=(x2-7x+10)(x2-7x-10)
Đặt t=x2-7x+10 đc:
B=t(t-20)=t2-20t
=t2-20t+100-100
=(t-10)2-100
Thay t=x2-7x+10 ta đc:
\(B=\left(x^2-7x+10-10\right)-100\ge0-100=-100\)
\(\Rightarrow B\ge-100\)
Dấu = khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)
Vậy MinB=-100 khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)
\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)
`a,4x-10=0 `
`<=> 4x=10`
`<=>x=10/4`
`<=>x=5/2`
`b, 7-3x=9-x `
`<=>-3x+x=9-7`
`<=>-2x=2`
`<=>x=-1`
`c, 2x-(3-5x) = 4(x+3)`
`<=>2x-3+5x=4x+12`
`<=>2x+5x-4x=12+3`
`<=>3x=15`
`<=>x=5`
`d, 5-(6-x)=4(3-2x) `
`<=>5-6+x=12-8x`
`<=>x+8x=12-5+6`
`<=>9x=13`
`<=>x=13/9`
`e, 4(x+3)=-7x+17 `
`<=>4x+12=-7x+17`
`<=>4x+7x=17-12`
`<=>11x=5`
`<=>x=5/11`
`f, 5(x-3) - 4=2(x-1)+7`
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`g, 5(x-3)-4=2(x-1)+7 `
`<=>5x-15-4=2x-2+7`
`<=>5x-2x=15+4-2+7`
`<=>3x=24`
`<=>x=8`
`h,4(3x-2)-3(x-4)=7x+20`
`<=>12x-8-3x+12=7x+20`
`<=>12x-3x-7x=20+8+12`
`<=>2x=40`
`<=>x=20`
\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
a) 5.(x^2-3x+1)+x.(1-5x)=x-2
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-14x-x=-2-5\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=\frac{7}{15}\)
b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)
\(\Leftrightarrow-4x^2+15x-10=0\)
Đề sai???
\(c,12x^2-4x\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)
\(\Leftrightarrow10x=-17\)
\(\Leftrightarrow x=-\frac{17}{10}\)
\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{3}{2}\)