f.2x^2+15x+7
g.3x^2-4x-7
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DH
19 tháng 7 2017
Làm 2 câu các câu còn lại tương tự!
a, \(E=-x^2+4x-5=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-2x-2x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2+1\ge1\Rightarrow-\left[\left(x-2\right)^2+1\right]\le-1\)
Hay \(E\le-1\) với mọi giá trị của \(x\in R\).
Để \(E=-1\) thì \(-\left[\left(x-2\right)^2+1\right]=-1\)
\(\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy.............
b, \(F=-2x^2+2x-1=-\left(2x^2-2x+1\right)\)
\(=-\left(2x^2-x-x+\dfrac{1}{2}-\dfrac{3}{2}\right)\)
\(=-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(2x-1\right)^2-\dfrac{3}{2}\ge-\dfrac{3}{2}\Rightarrow-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\le\dfrac{3}{2}\)
Hay \(F\le\dfrac{3}{2}\) với mọi giá trị của \(x\in R\).
Để \(F=\dfrac{3}{2}\) thì \(-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]=\dfrac{3}{2}\)
\(\Rightarrow\left(2x-1\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
20 tháng 7 2017
7, \(G=-4x^2+12x-7\)
\(=-4\left(x^2-3x+\dfrac{7}{4}\right)\)
\(=-4\left(x^2-\dfrac{3}{2}.x.2+\dfrac{9}{4}-\dfrac{2}{4}\right)\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+2\le2\)
Dấu " = " khi \(-4\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_G=2\) khi \(x=\dfrac{3}{2}\)
8, \(H=-2x^2+4x-15\)
\(=-2\left(x^2-2x+\dfrac{15}{2}\right)\)
\(=-2\left(x^2-2x+1+\dfrac{13}{2}\right)\)
\(=-2\left(x-1\right)^2-13\le-13\)
Dấu " = " khi \(-2\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_H=-13\) khi x = 1
9, \(K=-x^4+2x^2-2\)
\(=-\left(x^2-2x^2+1+1\right)\)
\(=-\left(x^2-1\right)^2-1\le-1\)
Dấu " = " khi \(-\left(x^2-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(MAX_K=-1\) khi \(x=\pm1\)
10, \(J=-3x^2+15x-9\)
\(=-3\left(x^2-\dfrac{5}{2}.x.2+\dfrac{10}{4}+\dfrac{2}{4}\right)\)
\(=-3\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{2}\le\dfrac{-3}{2}\)
Dấu " = " khi \(-3\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MAX_J=\dfrac{-3}{2}\) khi \(x=\dfrac{5}{2}\)
11 tháng 3 2023
`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`
TN
23 tháng 7 2017
a, \(A=9x^2-6x+5\)
\(=\left(9x^2-6x+1\right)+4\)
\(=\left(3x-1\right)^2+4\)
ta có:
\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)
Vậy Min A = 4
Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(b,B=4x^2-5x\)
\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)
\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)
TA có:
\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)
Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)
\(c,C=3x^2-6x\)
\(=3\left(x^2-2x+1\right)-3\)
\(=3\left(x-1\right)^2-3\)
Ta có:
\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)
vậy Min C = -3
Để C = -3 thì x-1=0 => x = 1
\(d,D=5x^2-15x\)
\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)
\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)
Ta có:
\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)
Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(e,E=x^2+3x+4\)
\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(f,F=2x^2-4x+7\)
\(=2\left(x^2-2x+1\right)+5\)
\(=2\left(x-1\right)^2+5\ge5\forall x\)
Vậy Min F = 5 khi x - 1 =0 => x = 1
\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)
\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)
Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
3 tháng 7 2015
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
f: =2x^2+x+14x+7
=x(2x+1)+7(2x+1)
=(2x+1)(x+7)
g: =3x^2+3x-7x-7
=3x(x+1)-7(x+1)
=(x+1)(3x-7)
\(f,2x^2+15x+7\\ =2x^2+x+14x+7\\=x\left(2x+1\right)+7\left(2x+1\right)\\ =\left(x+7\right)\left(2x+1\right)\\ g,3x^2-4x-7\\ =3x^2+3x-7x-7\\ =3x\left(x+1\right)-7\left(x+1\right)\\ =\left(3x-7\right)\left(x+1\right)\)