|a| = 2 a + x = 5a - x = 2 a - x =...
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|a| = 2 a + x = 5
a - x = 2 a - x = b
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NN
1
H9
HT.Phong (9A5)
CTVHS
31 tháng 10 2023
\(x^4-3x+2=\left(x-1\right)\left(x^3+bx^2+ax-2\right)\)
\(\Leftrightarrow x^4-3x+2=x^4+bx^3+ax^2-2x-x^3-bx^2-ax+2\)
\(\Leftrightarrow x^4-3x+2=x^4+\left(b-1\right)x^3+\left(a-b\right)x^2+\left(-2-a\right)x+2\)
\(\Leftrightarrow x^4+0x^3+0x^2-3x+2=x^4+\left(b-1\right)x^3+\left(a-b\right)x^2+\left(-2-a\right)x+2\)
Ta có: \(\left\{{}\begin{matrix}b-1=0\\a-b=0\\-2-a=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=1\\a=b\\a=1\end{matrix}\right.\)
\(\Leftrightarrow a=b=1\)
Vậy: ...
DT
2
18 tháng 1 2021
Ta có: \(f\left(x\right)=\dfrac{x-5}{3}\)
\(\Leftrightarrow f\left(x+2\right)=\dfrac{x+2-5}{3}=\dfrac{x-3}{3}\)
\(\Leftrightarrow f\left(x+2\right)=\dfrac{1}{3}\cdot\left(x-3\right)+0=\dfrac{1}{3}x-1\)
mà f(x+2)=ax+b
nên \(a=\dfrac{1}{3}\) và b=-1
hay \(a+b=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)
Vậy: \(a+b=-\dfrac{2}{3}\)
TT
1
5 tháng 11 2021
\(A⋮B\Leftrightarrow3x^3-2x^2+ax-a-5=\left(x-2\right)\cdot a\left(x\right)\)
Thay \(x=2\Leftrightarrow3\cdot8-2\cdot4+2a-a-5=0\)
\(\Leftrightarrow24-16-5+a=0\\ \Leftrightarrow a=-3\)
LH
1
HK
23 tháng 2 2022
\(a^2+b^2+3>ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+3\right)>2\left(ab+a+b\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(a^2-2ab+b^2\right)+4>0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b\right)^2+4>0\) \(\forall a,b\)
Vậy \(a^2+b^2+3>ab+a+b\forall a,b\)
NG
1
N
5
28 tháng 11 2016
\(\sqrt{x^2+a^2}=t>=IaI\); t^2=x^2+a^2
\(t-\frac{5a}{t}=x\) TH1. (x>=0 (*)
\(t^2-10a+\frac{25a^2}{t^2}=x^2=t^2-a^2\)
\(25a^2\cdot\left(\frac{1}{t}\right)^2+a^2-10a=0\)
\(t^2=\frac{25a^2}{10a-a^2}=0\)
\(x^2=\frac{25a}{\left(10-a\right)}-a^2\)
sau do Bien luan theo dk ton tai nghiem
x>=0; t>=IaI
TH2. x<0 (*) doi dau lai
Dai qua moi roi
2 tháng 10 2020
Bài 1:
a) \(\frac{x-1}{0-2}=\frac{1,2}{1,5}\)
\(\Leftrightarrow\frac{1-x}{2}=\frac{4}{5}\)
\(\Leftrightarrow5-5x=8\)
\(\Leftrightarrow x=-\frac{3}{5}\)
b) Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y+2z}{4-6+6}=\frac{16}{4}=4\)
\(\Rightarrow\hept{\begin{cases}x=4\\y=8\\z=12\end{cases}}\)
2 tháng 10 2020
Bài 1:
c) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)
\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)
d) \(x:y:z=3:5:2\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{5x-7y+5z}{15-35+10}=\frac{124}{-10}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{186}{5}\\y=-62\\z=-\frac{124}{5}\end{cases}}\)
9 tháng 1 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm a\end{matrix}\right.\)
Ta có: \(\dfrac{4}{x+a}+\dfrac{8}{x-a}=\dfrac{5a^2}{x\left(x^2-a^2\right)}\)
\(\Leftrightarrow\dfrac{4x\left(x-a\right)}{x\left(x+a\right)\left(x-a\right)}+\dfrac{8x\left(x+a\right)}{x\left(x+a\right)\left(x-a\right)}=\dfrac{5a^2}{x\left(x-a\right)\left(x+a\right)}\)
Suy ra: \(4x^2-4xa+8x^2+8xa-5a^2=0\)
\(\Leftrightarrow12x^2+4xa-5a^2=0\)
\(\Leftrightarrow12x^2+10xa-6xa-5a^2=0\)
\(\Leftrightarrow2x\left(6x+5a\right)-a\left(6x+5a\right)=0\)
\(\Leftrightarrow\left(6x+5a\right)\left(2x-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+5a=0\\2x-a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-5a\\2x=a\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5a}{6}\\x=\dfrac{a}{2}\end{matrix}\right.\)
|a| = 2 a + x = 5
a = -2 hoặc a = 2 a = 5 -x
a - x = 2 a - x = b
a = 2+x a = b+x