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a) \(250 + 250 × 2 = 250 + 500 = 750 \\ (250 + 250) × 2 = 500 × 2 = 1 000\)
b)\( 750 – 50 × 5 = 750 – 250 = 500\\
(750 – 50) × 5 = 700 × 5 = 3 500\)
c) \(210 × 4 – 4 × 210 = 840 – 840 = 0\\
3 × (270 : 9) × 0 = 0\)
Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
Bài 1 :
a) \(M=\dfrac{1}{2}x^2y.\left(-4\right)y\)
\(\Rightarrow M=-2x^2y^2\)
Khi \(x=\sqrt[]{2};y=\sqrt[]{3}\)
\(\Rightarrow M=-2.\left(\sqrt[]{2}\right)^2.\left(\sqrt[]{3}\right)^2\)
\(\Rightarrow M=-2.2.3=-12\)
b) \(N=xy.\sqrt[]{5x^2}\)
\(\Rightarrow N=xy.\left|x\right|\sqrt[]{5}\)
\(\Rightarrow\left[{}\begin{matrix}N=xy.x\sqrt[]{5}\left(x\ge0\right)\\N=xy.\left(-x\right)\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}N=x^2y\sqrt[]{5}\left(x\ge0\right)\\N=-x^2y\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
Khi \(x=-2< 0;y=\sqrt[]{5}\)
\(\Rightarrow N=-x^2y\sqrt[]{5}=-\left(-2\right)^2.\sqrt[]{5}.\sqrt[]{5}=-4.5=-20\)
2:
Tổng của 4 đơn thức là;
\(A=11x^2y^3+\dfrac{10}{7}x^2y^3-\dfrac{3}{7}x^2y^3-12x^2y^3=0\)
=>Khi x=-6 và y=15 thì A=0
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
\(A=\left|x-101\right|-101\)
\(\left|x-101\right|\ge0\)
\(\Rightarrow\left|x-101\right|-101\ge-101\)
\(\Rightarrow A\ge101\)
\(\Rightarrow MIN_A=101\Leftrightarrow\left|x-101\right|=0\)
\(\Rightarrow x=101\)
vay_
1:
a: \(A=2+3\sqrt{x^2+1}>=3\cdot1+2=5\)
Dấu = xảy ra khi x=0
b: \(B=\sqrt{x+8}-7>=-7\)
Dấu = xảy ra khi x=-8
a) \(A=1-8x-x^2=-\left(x^2+8x+16\right)+17=-\left(x-4\right)^2+17\le17\)
\(ĐTXR\Leftrightarrow x=4\)
b) \(B=5-2x+x^2=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
\(ĐTXR\Leftrightarrow x=1\)
c) \(C=x^2+4y^2-6x+8y-2021=\left(x^2-6y+9\right)+\left(4y^2+8y+4\right)-2034=\left(x-3\right)^2+\left(2y+2\right)^2-2034\ge-2034\)
\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
a: Ta có: \(A=-x^2-8x+1\)
\(=-\left(x^2+8x-1\right)\)
\(=-\left(x^2+8x+16-17\right)\)
\(=-\left(x+4\right)^2+17\le17\forall x\)
Dấu '=' xảy ra khi x=-4
b: Ta có: \(x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
a)
948 – 429 + 479
= 519 + 479
= 998
424 : 2 x 3
= 212 x 3
= 636
b)
750 – 101 x 6
= 750 – 606
= 144
100 : 2 : 5
= 50 : 5
= 10
c)
998 – (302 + 685)
= 998 – 987
= 11
(421 – 19) x 2
= 402 x 2
= 804