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Bài 1:
\(A=-x^2-2x+9\)
\(A=-\left(x^2+2x-9\right)\)
\(A=-\left(x^2+2x+1-10\right)\)
\(A=-\left(x+1\right)^2+10\)
Vì \(-\left(x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+1\right)^2+10\le10\)
\(\Rightarrow Amax=10\Leftrightarrow x=-1\)
\(B=-9x^2+6x+25\)
\(B=-\left(9x^2-6x-25\right)\)
\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)
\(B=-\left(3x-1\right)^2+26\)
Vì \(-\left(3x-1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(3x-1\right)^2+26\le26\)
\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(C=-x^2+x+1\)
\(C=-\left(x^2-x-1\right)\)
\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)
\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)
Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(D=-2x^2+3x+1\)
\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)
\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)
\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)
Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x
\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)
\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(E=-25x^2-10x+7\)
\(E=-\left(25x^2+10x-7\right)\)
\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)
\(E=-\left(5x+1\right)^2+8\)
Vì \(-\left(5x+1\right)^2\le0\) với mọi x
\(\Rightarrow-\left(5x+1\right)^2+8\le8\)
\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
Bài 2:
\(A=9x^2+6x+4\)
\(A=\left(3x\right)^2+2.3x+1+3\)
\(A=\left(3x+1\right)^2+3\)
Vì \(\left(3x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(3x+1\right)^2+3\ge3\)
\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)
\(B=4x^2+4x+12\)
\(B=\left(2x\right)^2+2.2x+1+11\)
\(B=\left(2x+1\right)^2+11\)
Vì \(\left(2x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+1\right)^2+11\ge11\)
\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)
\(C=x^2+x+3\)
\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)
\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
\(D=2x^2+3x+1\)
\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)
\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)
\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)
Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)
\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)
\(E=64x^2+16x+3\)
\(E=\left(8x\right)^2+2.8x+1+2\)
\(E=\left(8x+1\right)^2+2\)
Vì \(\left(8x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(8x+1\right)^2+2\ge2\)
\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)
\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
(6x+1)(2x-5)=12x2-30x+2x-5=12x2-28x-5
(2x+5)2-2x(2x+8)=4x2+20x+25-4x2-16x=4x+25
(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x2-3x-10x+5=6x2-13x+5
(X-1)2-(x+1)(x-1)=x2-2x+1-x2+1=-2x+2
(3x+2)(9x2-6x+4)-(3+x)(x-3)=27x3+8+9-x2=27x3-x2+17
A = ( 3x )3 + 23 - 27x3 + 6 = 27x3 + 8 - 27x3 + 6 = 14 ( đpcm )
B = x3 + 3x2 + 3x + 1 - ( x3 - 1 ) - 3x2 - 3x = x3 + 1 - x3 + 1 = 2 ( đpcm )
C = 6( x + 2 )( x2 - 2x )( x2 - 2x + 4 ) - 6x3 - 2 ( bạn xem lại đề bài nhé ._. )
D = 2[ ( 3x )3 + 13 ] - 54x3 = 2( 27x3 + 1 ) - 54x3 = 54x3 + 2 - 54x3 = 2 ( đpcm )
Bài 1:
a: A=x^2-6x+10
=x^2-6x+9+1
=(x-3)^2+1>=1
Dấu = xảy ra khi x=3
b: \(B=3x^2-12x+1\)
=3(x^2-4x+1/3)
=3(x^2-4x+4-11/3)
=3(x-2)^2-11>=-11
Dấu = xảy ra khi x=2
a) Ta có: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0\)
\(\Leftrightarrow2x^2+9x-5-2x^2+x+3=0\)
\(\Leftrightarrow10x-2=0\)
hay 10x=2
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy: \(x=\frac{1}{5}\)
b) Ta có: \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\)
\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\Leftrightarrow2x-6=0\)
hay 2x=6
\(\Leftrightarrow x=3\)
Vậy: x=3
c) Ta có: \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
hay \(x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)
\(\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\)
\(\Leftrightarrow3x^2-x-10=2x^2-2x-4\)
\(\Leftrightarrow3x^2-x-10-2x^2+2x+4=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
đ) Ta có: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
e) Ta có: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
a) $(x+5)(2x-1)=(2x-3)(x+1)$
$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$
$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$
b)
$(x+1)(x+9)=(x+3)(x+5)$
$\Leftrightarrow x^2+10x+9=x^2+8x+15$
$\Leftrightarrow 2x=6\Rightarrow x=3$
c)
$(3x+5)(2x+1)=(6x-2)(x-3)$
$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$
$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^