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a) <=> |-5X| =3X +16
DK : X >-16/3
-5X = 3X +16 HOAC -5X =-3X-16
-8X = 16 HOAC -2X = -16
X= -2 HOAC X= 8
VẬY S= {-2; 8}
b) <=> 3X +X = 1+2
<=> 4X = 3
<=> X=3/4
VẬY S={3/4}
c) DK : X> 10/4
-2X = 4X-10 HOAC -2X = -4X +10
-6X = 10 HOAC 2X = 10
X= -5/3 (LOAI) HOAC X= 5 (NHAN)
VẬY S={5}
LƯU Ý: CÓ CHỮ " HOẶC" THÌ KHÔNG CẦN MŨI TÊN HAI CHIỀU
-MÌNH CHỈ GHI CÁCH GIẢI THÔI NHÉ
CHÚC BẠN HỌC TỐT .
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`
`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`
`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`
`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`
`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`
`\Leftrightarrow 10x^2 - 19x = 0`
`\Leftrightarrow x(10x - 19)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)
Vậy, `x={0; 19/10}.`
c) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
Bằng cách nhân chéo ta đc đẳng thức :
\(\frac{5x-1}{3x+2}=\frac{5x-1}{2x+1}\)
=> 3x + 2 = 2x + 1
=> 3x - 2x = 1 - 2
=> x = -1
Vậy,..........
\(\left(5x-1\right)\left(3x+2\right)=\left(5x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(5x-1\right)\left(3x+2\right)-\left(5x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(3x+2-2x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=-1\end{cases}}}\)
=.= hok tốt!!
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
d) \(2x^3+3x^2+3x+1=2x^3+x^2+2x^2+x+2x+1\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2-x+1\right)\)
a) Ta có: \(|-5x|-16=3x\)
Đk: \(3x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\orbr{\begin{cases}-5x-16=3x\\5x-16=3x\end{cases}}\Rightarrow\orbr{\begin{cases}-5x-3x=16\\5x-3x=16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-8x=16\\-2x=16\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
Mà x \(\ge0\)\(\Rightarrow x=8\)
b) \(|3x-2|=1-x\)
\(\Rightarrow\orbr{\begin{cases}3x-2=1-x\\3x-2=-1+x\end{cases}\Rightarrow}\orbr{\begin{cases}3x+x=1+2\\3x-x=-1+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=3\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\end{cases}}\)
Vậy: x = \(\frac{3}{4}\)hoặc x\(=\frac{1}{2}\)
c) Ta có: \(|-2x|=4x-10\)
Đk: \(4x-10\ge0\Rightarrow4x\ge10\Rightarrow x\ge\frac{5}{2}\)
\(\Rightarrow\orbr{\begin{cases}-2x=4x-10\\2x=4x-10\end{cases}}\Rightarrow\orbr{\begin{cases}-2x-4x=-10\\2x-4x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-6x=-10\\-2x=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=5\end{cases}}\)
mà x\(\ge\frac{5}{2}\)\(\Rightarrow x=5\)