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\(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Min A = 10 khi: 2x + 1 = 0
<=> x = -1/2
\(4.\)
\(a.A=5-8x-x^2\)
\(=-\left(16+8x+x^2\right)+21\)
\(=-\left(4+x\right)^2+21\le21\)
\(A_{max}=21\)
Dấu '='xảy ra khi \(x=-4\)
\(b.B=5-x^2+2x-4y^2-4y\)
\(=-\left(1-2x+x^2\right)-\left(4+4y+4y^2\right)+10\)
\(=-\left(1-x\right)^2-\left(2+2y\right)^2+10\le10\)
\(B_{max}=10\)
Dấu '=' xảy ra khi \(x=1;y=-1\)
\(5.\)
\(a.\) Ta có:\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\left(1\right)\)
hay\(b-c=0\Leftrightarrow b=c\left(2\right)\)
hay\(c-a=0\Leftrightarrow c=a\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\)và\(\left(3\right)\)suy ra:\(a=b=c\left(đpcm\right)\)
\(b.a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)
hay\(b+2=0\Leftrightarrow b=-2\)
hay\(2c-2=0\Leftrightarrow c=1\)
V...
^^
a) \(A=5-8x-x^2\)
\(=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+2.x.4+4^2-16-5\right)\)
\(=-\left[\left(x+4\right)^2-21\right]\)
\(=-\left(x+4\right)^2+21\le21\)
Dấu "=" khi x + 4 = 0 => x = -4
Vậy GTLN của A là 21 khi x = -4
b) \(B=5-x^2+2x-4y^2-4y\)
\(=-\left(x^2-2x+4y^2+4y-5\right)\)
\(=-\left[x^2-2x+1+\left(2y\right)^2+2.2y.1+1-7\right]\)
\(=-\left[\left(x-1\right)^2+\left(2y+1\right)^2\right]+7\le7\)
Dấu "=" khi \(\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=\frac{-1}{2}\end{cases}}}\)
Vậy GTLN của B là 7 khi x = 1 và y = -1/2
c) Theo đề: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)(ĐPCM)
d) \(a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(\text{4c^2}-4c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a-1=0\\b+2=0\\2c-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}}\)
Vậy nghiệm phương trình: a = 1; b = -2; c = 1/2
Chúc bạn học tốt ^_^
a>(8x^2y+10xy6^2-6xy):2xy=4xy+5y-3
b>(3x^2-4x).(2x-6)=6x^3-26x^2+24x
Ta có : A = x2 - 4x + 1
=> A = x2 - 2.x.2 + 4 - 3
=> A = (x - 2)2 - 3
Mà : (x - 2)2 \(\ge0\forall x\in R\)
Nên : (x - 2)2 - 3 \(\ge-3\forall x\in R\)
Vậy GTNN của A là -3 khi x = 2
\(B=4x^2+4x+11=\left(2x\right)^2+2.2x.1+1+10=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\Rightarrow B=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi (2x+1)2=0 <=> 2x+1=0 <=> x=-1/2
Vậy gtnn của B là 10 khi x=-1/2
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\(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi x=0 hoặc x=-5
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
^^
1) a) \(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(99+98\right)+....\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+.....+2+1\)
\(=\dfrac{100.101}{2}=5050\)
2) a) \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3a^2b+3ab^2-3a^2b+3ab^2=a^3+b^3=VT\)
b) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b+3ab^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)Khi \(a^3+b^3+c^3=3abc\) \(\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
i.i \(A=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{3}{abc}=3\)iii. \(a^3+b^3+c^3=3abc\Rightarrow\)
\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: a=b=c
\(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
TH2: a+b+c=0
\(B=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
=>-(x2+8x-5)
=>-(x2 +2.4x+16-5-16)
=>-[(x+4)2-21]
=>-(x+4)2+21
do -(x+4)2 luôn > hoặc= 0 với mọi x
=>-(x+4)2+21 >/21
=>A>/21
=>GTLN của A=21 Khi x+4=0
x=-4
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