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\(A=a-2ab+6b\)
thay \(|a|=1,5\)và \(b=-0,5\)vào biểu thức
\(A=1,5-2.1,5.\left(-0,5\right)+6.\left(-0,5\right)\)
\(A=1.1,5-2.1,5.\left(-0,5\right).\left(1+6\right)\)
\(A=1,5.\left(1-2\right).\left(0,5\right).\left(7\right)\)
\(A=1,5.\left(-1\right).0,5.\left(7\right)\)
\(A=\frac{3}{2}.\left(-1\right).\frac{-1}{2}.7\)
\(A=\frac{-3}{2}.\frac{-7}{2}\)
\(A=\frac{-21}{4}\)
\(\frac{x}{y}=a\Rightarrow x=ay\)
\(\Rightarrow\frac{x+y}{x-y}=\frac{ay+y}{ay-y}=\frac{y\left(a+1\right)}{y\left(a-1\right)}=\frac{a+1}{a-1}\)
\(\frac{a}{b}=2\Rightarrow a=2b;\frac{c}{b}=3\Rightarrow c=3b\Rightarrow c-b=2b\)
\(\Rightarrow a=c-b\)
\(\Rightarrow\frac{a+c}{b+c}=\frac{c-b+b}{b+c}=\frac{b}{b+c}\)
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\Rightarrow c=\frac{2ab}{a+b}\)
\(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{\frac{a^2+ab-2ab}{a+b}}{\frac{2ab-ab-b^2}{a+b}}=\frac{a^2+ab-2ab}{2ab-ab-b^2}=\frac{a.\left(a-b\right)}{b.\left(a-b\right)}=\frac{a}{b}\)(ĐPCM)
\(\left|2x-27\right|^{2017}+\left(3y+10\right)^{2012}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)(làm tắt nha, có gì bn thêm vào)
câu 2 : | 2x - 27 |\(^{2011}\)+ ( 3y + 10 ) \(^{2012}\)=0
=> \(\left|2x-27\right|^{2011}\)lớn hơn hoặc = 0 (1)
=> \(\left(3y+10\right)^{2012}\)>hoặc = 0(2)
mà (1) + (2) =0
nên => \(\left|2x-27\right|^{2011}=0\)và \(\left(3y+10\right)^{2012}=0\)
\(\left|2x-27\right|^{2011}=0^{2011}\) \(\left(3y+10\right)^{2012}=0^{2012}\)
\(\left|2x-27\right|=0\) 3y + 10 = 0
2x = 27 3y = -10
x = 27 : 2 y = -10 : 3
x = 13,5 y = \(\frac{-10}{3}\)
áp dụng t/c DTSBN,ta có:
\(\frac{ab+ac}{2}=\frac{bc+ab}{3}=\frac{ca+bc}{4}=\frac{ab+ac-bc-ab+ca+bc}{2-3+4}=\frac{2ac}{3}\)
\(\frac{ab+ac}{2}=\frac{2ac}{3}\Leftrightarrow3ab+3ac=4ac\Leftrightarrow3ab=ac\Leftrightarrow3b=c\Leftrightarrow\frac{b}{1}=\frac{c}{3}\Rightarrow\frac{b}{5}=\frac{c}{15}\)(vì a khác 0)(!)
\(\frac{ca+cb}{4}=\frac{2ac}{3}\Leftrightarrow3ac+3cb=8ac\Leftrightarrow3bc=5ac\Rightarrow3b=5a\Rightarrow\frac{a}{3}=\frac{b}{5}\)(vì c khác 0)(@)
từ (!) và (@) => đpcm
a) sai đề rồi bn
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\left(\frac{a+b}{c+d}\right)^3\)(tính chất dãy tỉ số bằng nhau) (1)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3-b^3}{c^3-d^3}\)(2)
từ (1) và (2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\left(đpcm\right)\)
bài 1 :
a, A = 3|2x - 1| - 5 = 0
có 3|2x - 1| > 0
=> A > -5
xét A = -5 khi
|2x - 1| = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
vậy Min A = -5 khi x = 1/2
b, c, d, làm tương tự
Bài 1:
\(a)A=3|2x-1|-5\)
Vì \(|2x-1|\ge0\)\(\forall x\)
\(\Rightarrow3|2x-1|\ge0\) \(\forall x\)
\(\Rightarrow3|2x-1|-5\ge-5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-5\Leftrightarrow x=\frac{1}{2}\)
\(b)x^2+3|y-2|-1\)
Vì \(\hept{\begin{cases}x^2\ge0\forall x\\3|y-2|\ge0\forall y\end{cases}}\)
\(\Rightarrow x^2+3|y-2|-1\ge-1\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x^2=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy \(Min_B=-1\Leftrightarrow x=0,y=2\)
\(c)\left(2x^2+1\right)^4-3\)
Vì \(\left(2x^2+1\right)^4\ge0\)\(\forall x\)
\(\Rightarrow\left(2x^2+1\right)^4-3\ge-3\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x^2+1=0\)
\(\Leftrightarrow2x^2=-1\)
\(\Leftrightarrow x^2=-\frac{1}{2}\left(voli\right)\)
Vậy không tìm được gt x
\(d)D=|x-\frac{1}{2}|+\left(y+2\right)^2+11\)
Vì \(\hept{\begin{cases}|x-\frac{1}{2}|\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\)
\(\Rightarrow|x-\frac{1}{2}|+\left(y+2\right)^2+11\ge11\) \(\forall x,y\)
Dấu '=' xảy ra:
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-2\end{cases}}\)
Vậy \(Min_D=11\Leftrightarrow x=\frac{1}{2},y=-2\)
Bài 2:
\(a)A=10-5|x-2|\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow5|x-2|\ge0\)\(\forall x\)
\(\Rightarrow\)\(10-5|x-2|\le10\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_A=10\Leftrightarrow x=2\)
\(b)B=5-|2x-1|^2\)
Vì \(|2x-1|^2\ge0\)\(\forall x\)
\(\Rightarrow5-|2x-1|^2\le5\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Max_B=5\Leftrightarrow x=\frac{1}{2}\)
\(c)C=\frac{1}{|x-2|+3}\)
Vì \(|x-2|\ge0\)\(\forall x\)
\(\Rightarrow|x-2|+3\ge3\) \(\forall x\)
\(\Rightarrow\frac{1}{|x-2|+3}\le\frac{1}{3}\) \(\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(Max_C=\frac{1}{3}\Leftrightarrow x=2\)
1. \(AB=-\frac{1}{3}x^2y^2\cdot\left(-6x^3y^4\right)=\left(-\frac{1}{3}\cdot-6\right)\left(x^2x^3\right)\left(y^2y^4\right)=2x^5y^6\)
Bậc = 5 + 6 = 11
2. Thiếu B