a)\(\left|5x+4\right|+7=26\)
b)\(3\left|9-2x\right|-17=16\)
c)\(8x-\left|4x+1\right|=x+2\)
d)\(\left|17x-5\right|-\left|17x+5\right|=0\)
e)\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|=2\)
Tìm X
Những câu hỏi liên quan
Tìm x biết1) left(2x+3right)left(x-4right)+left(x-5right)left(x-2right)left(3x-5right)left(x-4right)2)left(8x-3right)left(3x+2right)-left(4x+7right)left(x+4right)left(2x+1right)left(5x+1right)-333)6xleft(3x+5right)-2xleft(9x-2right)+left(17-xright)left(x-1right)+xleft(x-18right)-17x^204)left(x-1right)left(x+2right)-left(x-3right)+5x-70Giúp mình nha. Camon nhiều
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Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
tìm x biết
a.\(\left||x+5\right|-4|=3\)
b.8x-\(\left|4x+1\right|\)=x+2
c.\(\left|17x-5\right|-\left|17x+5\right|=0\)
d.\(\left|x-1\right|=2x-5\)
Tìm số nguyên x
1/ 17x+3left(-16x-37right)2x+43-4x
2/ -2x-3left(x-17right)34-2left(-x+25right)
3/ left{-3x+2left[45-x-3left(3x+7right)-2xright]+4xright}55-103-57:left[-2left(2x-1right)^2-left(-9right)^0right]-106
4/ -2x+3left{12-2left[3x-left(20+2xright)-4xright]+1right}45
5/ 3x-32-5+1
6/ 15+4x 2x-145
7/ -3left(2x+5right)-16 -4left(3-2xright)
8/ -2x+15 3x-7 19-x
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Tìm số nguyên x
1/ \(17x+3\left(-16x-37\right)=2x+43-4x\)
2/ \(-2x-3\left(x-17\right)=34-2\left(-x+25\right)\)
3/ \(\left\{-3x+2\left[45-x-3\left(3x+7\right)-2x\right]+4x\right\}=55-103-57:\left[-2\left(2x-1\right)^2-\left(-9\right)^0\right]=-106\)
4/ \(-2x+3\left\{12-2\left[3x-\left(20+2x\right)-4x\right]+1\right\}=45\)
5/ \(3x-32>-5+1\)
6/ \(15+4x< 2x-145\)
7/ \(-3\left(2x+5\right)-16< -4\left(3-2x\right)\)
8/ \(-2x+15< 3x-7< 19-x\)
bài tập tết nâng cao phải ko
mk cũng có nhưng chưa làm dc
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Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
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b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
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c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha
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Tìm x :
a) \(4x^2-8x=0\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
c) \(2x\left(x-4\right)^2-\left(x-4\right)=0\)
d) \(3x\left(x-1\right)+\left(1-x\right)^2=0\)
e) \(5x\left(x-9\right)^2-\left(9-x\right)^3=0\)
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
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\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
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a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy...
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(1-3x\right)=0\)
\(\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy..
d) \(3x\left(x-1\right)+\left(1-x\right)^2=0\)
\(\Rightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(3x-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy..
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Xem thêm câu trả lời
Bài 2 : Giải các bất phương trình sau :
11 , left(2x-7right)left(4-5xright)ge0
12 , x^2-x-202left(x-11right)
13 , 3xleft(2x+7right)left(9-3xright)ge0
14 , x^3+8x^2+17x+10 0
15 , x^3+6x^2+11x+60
16 , frac{left(2x-5right)left(x+2right)}{-4x+3}0
17 , frac{x-3}{x+1}frac{x+5}{x-2}
18 , frac{x-3}{x+5} frac{1-2x}{x-3}
19 , frac{3x-4}{x-2}1
20 , frac{2x-5}{2-x}ge-1
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Bài 2 : Giải các bất phương trình sau :
11 , \(\left(2x-7\right)\left(4-5x\right)\ge0\)
12 , \(x^2-x-20>2\left(x-11\right)\)
13 , \(3x\left(2x+7\right)\left(9-3x\right)\ge0\)
14 , \(x^3+8x^2+17x+10< 0\)
15 , \(x^3+6x^2+11x+6>0\)
16 , \(\frac{\left(2x-5\right)\left(x+2\right)}{-4x+3}>0\)
17 , \(\frac{x-3}{x+1}>\frac{x+5}{x-2}\)
18 , \(\frac{x-3}{x+5}< \frac{1-2x}{x-3}\)
19 , \(\frac{3x-4}{x-2}>1\)
20 , \(\frac{2x-5}{2-x}\ge-1\)
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
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8 tháng 2 2022 lúc 21:50
Bài Tập: Giải phương trình :
a) (x + 5)(2x - 3) = 0
b) \(\left(x^2-9\right)\left(4-x\right)=0\)
c) \(\left(2x+3\right)\left(4-5x\right)=0\)
d) \(2\left(x+3\right)\left(x-4\right)=0\)
e) \(\left(x^2-9\right)\left(4-x\right)=0\)
f) \(\left(2x+3\right)\left(x^2-16\right)=0\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
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a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
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d) \(^{ }4x\left(2x+3\right)-8x\left(x+4\right)\)
e) \(^{ }2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
f) \(^{ }x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(4x\left(2x+3\right)-8x\left(x+4\right)\)
\(=8x^2+12x-8x^2-32x\)
=-20x
e: Ta có: \(2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\)
\(=10x^2+4x+6x^2-2x-9x+3\)
\(=16x^2-7x+3\)
f: Ta có: \(x\left(x+2\right)^2-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3+4x^2+4x-x^3-3x^2-3x-1+3x^2-3\)
\(=4x^2+x-4\)
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