Tìm x,y để các biểu thức sau đạt GTNN, tính giá trị đó:
a) \(M=x^2-8x+2018\)
b) \(N=4x^2-12x+2019\)
c) \(P=x^2-x+2016\)
d) \(Q=x^2-2x+y^2+4y+2020\)
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Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0
=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0
Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1
Thay vào bt S :
S = ( 2 - 1)^2019 + (2-1)^2019
= 1^2019 + 1^2019 = 2
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Bài 1:
a) Ta có: \(A=-x^2-4x-2\)
\(=-\left(x^2+4x+2\right)\)
\(=-\left(x^2+4x+4-2\right)\)
\(=-\left(x+2\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-2
b) Ta có: \(B=-2x^2-3x+5\)
\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)
c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)
\(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9\le9\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 2:
a) Ta có: \(=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)
b) Ta có: \(B=9x^2-6xy+2y^2+1\)
\(=9x^2-6xy+y^2+y^2+1\)
\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)
c) Ta có: \(E=x^2-2x+y^2-4y+6\)
\(=x^2-2x+1+y^2-4y+4+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)
a,\(x^2-6x-17=x^2-2\cdot3x+9-26=\left(x-3\right)^2-26\ge-26\)
b, \(x^2-10x=x^2-2\cdot5x+25-25=\left(x-5\right)^2-25\ge-25\)
c,\(3x^2-12x+5=3x^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+12-7=\left(\sqrt{3}x-2\sqrt{3}\right)^2-7\ge-7\)
d,\(2x^2-x-1=2x^2-2\cdot\sqrt{2}x\cdot\dfrac{1}{2\sqrt{2}}+\dfrac{1}{8}-\dfrac{9}{8}=\left(\sqrt{2}x-\dfrac{1}{2\sqrt{2}}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
e,\(x^2+y^2-8x+4y+27=x^2-2\cdot4x+16+y^2+2\cdot2y+4+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\)
f,\(x\left(x-6\right)=x^2-6x=x^2-2\cdot3x+9-9=\left(x-3\right)^2-9\ge-9\)
h,\(\left(x-2\right)\cdot\left(x-5\right)\cdot\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)=\left(x^2-7x\right)^2-100\ge-100\)
Mình giúp tính biểu thức thôi
còn lại bạn tự làm nhé
a: \(M=2\left[\left(a+b\right)^3-3ab\left(a+b\right)\right]-3\left[\left(a+b\right)^2-2ab\right]\)
\(=2\left(1-3ab\right)-3\left(1-2ab\right)\)
\(=2-6ab-3+6ab=-1\)
b: \(4x^4+2x^2+a⋮x-2\)
\(\Leftrightarrow4x^4-8x^3+8x^3-16x^2+14x^2-56+a+56⋮x-2\)
=>a+56=0
=>a=-56
c: \(A=x^2+8x+16+4y^2+4y+1-34\)
\(=\left(x+4\right)^2+\left(2y+1\right)^2-34>=-34\)
Dấu = xảy ra khi x=-4 và y=-1/2
d: \(\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+2\)
\(\Leftrightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+2\)
=>-4x^2-10x-8=-4x^2+2
=>-10x=10
=>x=-1
x^2-5x-3=0
\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(-3\right)=25+12=37\)>0
=>PT có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{5-\sqrt{37}}{2}\\x_2=\dfrac{5+\sqrt{37}}{2}\end{matrix}\right.\)
e: \(\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
Đề lỗi font. Bạn cần chỉnh sửa lại bằng công thức toán để được hỗ trợ tốt hơn.
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
a) \(M=x^2-8x+2018=x^2-8x+16+2002=\left(x-4\right)^2+2002\)
\(\left(x-4\right)^2\ge0\forall x\Rightarrow\left(x-4\right)^2+2002\ge2002\)
Dấu " = " xảy ra <=> x - 4 = 0 => x = 4
Vậy MMin = 2002 khi x = 4
b) \(N=4x^2-12x+2019=4x^2-12x+9+2010=\left(2x-3\right)^2+2010\)
\(\left(2x-3\right)^2\ge0\forall x\Rightarrow\left(2x-3\right)^2+2010\ge2010\)
Dấu " = " xảy ra <=> 2x - 3 = 0 => x = 3/2
Vậy NMin = 2010 khi x = 3/2
c) \(P=x^2-x+2016=x^2-x+\frac{1}{4}+\frac{8063}{4}=\left(x-\frac{1}{2}\right)^2+\frac{8063}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{8063}{4}\ge\frac{8063}{4}\)
Dấu " = " xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy PMin = 8063/4 khi x = 1/2
d) \(Q=x^2-2x+y^2+4y+2020\)
\(Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2015\)
\(Q=\left(x-1\right)^2+\left(y+2\right)^2+2015\)
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+2015\ge2015\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy QMin = 2015 khi x = 1 ; y = -2