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\(A=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)
\(A_{min}=10\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)
\(B=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
\(B_{min}=-36\) khi \(x^2+5x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(C=\left(x^2-2x+1\right)+\left(y^2-4x+4\right)+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left(x;y\right)=\left(1;2\right)\)
a/ \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)
\(=-3x^2+7x-4\)
Thay x = 2 vào A được:
\(=-3.2^2+7.2-4=-2\)
Vậy: Giá trị của A khi x = 2 là -2
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b/ \(B=126y^3+\left(x-5y\right)\left(x^2+25y^2+5xy\right)\)
\(=126y^3+x^3-125y^3\)
Thay x = -5 và y = -3 vào B được:
\(126.\left(-3\right)^3+\left(-5\right)^3-125.\left(-3\right)^3=-152\)
Vậy: Giá trị của B tại x = -5 và y = -3 là -152
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c/ \(C=a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)
\(=a^3+b^3-\left(a-b\right)^3\)
\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=2b^3+3a^2b-3ab^2\)
Thay a = -4 và b = 4 vào C được:
\(2.4^3+3.\left(-4\right)^2.4-3.\left(-4\right).4^2=512\)
Vậy: Giá trị của C tại a = -4 vào b = 4 là 512
a:Ta có: \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)
\(=-3x^2+7x-4\)
\(=-3\cdot2^2+7\cdot2-4\)
\(=-12-4+14=-2\)
c: Ta có: \(C=a^3+b^3-\left(a-b\right)\left(a^2-2ab+b^2\right)\)
\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=2b^3+3a^2b-3ab^2\)
\(=2\cdot4^3+3\cdot\left(-4\right)^2\cdot4-3\cdot\left(-4\right)\cdot4^2\)
\(=128+192+192=512\)
`@` `\text {Ans}`
`\downarrow`
`A= (2x - 3)^2 - (2x + 3)^2`
`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`
`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`
`= -6 * 4x`
`= -24x`
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
A=(x2-4x+4)-5=(x-2)2-5≥-5
Dau bang xay ra khi: x=2
Vay GTNN cua A=-5 khi x=2
B=(4x2+4x+1)+10=(2x+1)2+10≥10
Dau bang xay ra khi: x=-1/2
Vay GTNN cua B=10 khi x=-1/2
C=[(x-1)(x+6)].[(x+2)(x+3)]
= (x2+5x-6)(x2+5x+6)
Dat x2+5x=a => (a-6)(a+6)=a2-36≥-36
Dau bang xay ra khi : a=0 => x=0 hoac x=-5
Vay GTNN cua C=-36 khi x=0 hoac c=-5
D=-(x2+8x-5)
=> -D=x2+8x-5=(x2+8x+16)-21=(x+4)2-21
=> D= 21-(x+4)2≤21
Dau bang xay ra khi : x=-4
Vay GTLN cua D=21 khi x=-4
E=-(x2-4x-1)=-(x2-4x+4-5)=-(x-2)2+5=5-(x-2)2≤5
Dau bang xay ra khi : x=2
Vay GTLN cua E=5 khi x=2
\(A=x^2-4x+1\\ =x^2-4x+4-3\\ =\left(x^2-4x+4\right)-3\\ =\left(x-2\right)^2-3\\ \text{Do }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-2\right)^2-3\ge-3\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Min\right)}=-3\text{ }khi\text{ }x=2\)
\(B=4x^2+4x+11\\ =4x^2+4x+1+10\\ =\left(4x^2+4x+1\right)+10\\ =\left(2x+1\right)^2+10\\ \text{Do }\left(2x+1\right)^2\ge0\forall x\\ \Rightarrow B=\left(2x+1\right)^2+10\ge10\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(2x+1\right)^2=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\\ \\ \text{Vậy }B_{\left(Min\right)}=10\text{ }khi\text{ }x=-\dfrac{1}{2}\)
\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\\ =\left(x^2-x+6x-6\right)\left(x^2+3x+2x+6\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ =\left(x^2+5x\right)-36\\ \text{Do }\left(x^2+5x\right)^2\ge0\forall x\\ \Rightarrow C=\left(x^2+5x\right)^2-36\ge-36\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x^2+5x\right)^2=0\\ \Leftrightarrow x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ \text{Vậy }C_{\left(Min\right)}=-36\text{ }khi\text{ }x=-0\text{ hoặc }x=-5\)
a) Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3\left(x^2-1\right)\)
\(=4x-3x^2+3\)
\(=-3x^2+4x+3\)
b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)
\(=5x^2-20-32x^2+48x-16+17\)
\(=-27x^2+48x-19\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)
\(\Rightarrow x=29\)
2)
a) \(=x^2-9-x^2+6x-9=6x-18\)
b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)
Ta có : (x + 4)2 - (x + 1)(x - 1) = 16
<=> x2 + 8x + 16 - (x2 - 1) = 16
<=> x2 + 8x + 16 - x2 + 1 = 16
<=> 8x + 17 = 16
=> 8x = -1
=> x = \(-\frac{1}{8}\)
Ta có : x2 - 4x + 4 =0
<=> x2 - 2.x.2 + 22 = 0
<=> (x - 2)2 = 0
=> x - 2 = 0
=> x = 2