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a: =>|5x-2|=|2x-3|
=>5x-2=2x-3 hoặc 5x-2=-2x+3
=>3x=-1 hoặc 7x=5
=>x=5/7 hoặc x=-1/3
b: =>|5x-2|-|2x+2|=3x+5
TH1 x<-1
PT sẽ là 2-5x+2x+2=3x+5
=>-3x+4=3x+5
=>-6x=1
=>x=-1/6(loại)
TH2: -1<=x<2/5
Pt sẽ là 2-5x-2x-2=3x+5
=>-7x=3x+5
=>-4x=5
=>x=-5/4(loại)
Th3: x>=2/5
PT sẽ là 5x-2-2x-2=3x+5
=>3x-4=3x+5
=>0x=9(loại)
a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)
Vậy: \(S=\left\{-1\right\}\)
b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)
Vậy: \(S=\left\{15\right\}\)
c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)
Vậy: \(S=\left\{5\right\}\)
d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)
Vậy: \(S=\left\{12\right\}\)
e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
a)3x-2=2x-3
=>x=-1
b)7-2x=22-3x
=>x=15
c)8x-3=5x+12
=>3x=15
=>x=5
d)x-12+4x=25+2x-1
=>3x=12
=>x=4
e)x+2x+3x-19=3x+5
=>3x=24
=>x=8
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
Giải:
a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)
\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)
\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)
\(\Leftrightarrow30x-37=4\)
\(\Leftrightarrow30x=41\)
\(\Leftrightarrow x=\dfrac{41}{30}\)
Vậy ...
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)
\(\Leftrightarrow x^3+8-x^3-3=14x\)
\(\Leftrightarrow5=14x\)
\(\Leftrightarrow x=\dfrac{5}{14}\)
Vậy ...
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
\(\Leftrightarrow x^3+1-x^3-3x=2\)
\(\Leftrightarrow1-3x=2\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)
=> \(x^2-2x-x-3x+7+9x=6\)
=> \(x^2-2x-x^2-3x+7+9x=6\)
=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)
=> \(4x=-1\)
Vậy \(x=\dfrac{-1}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)
=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)
=> \(42x=43\)
Vậy \(x=\dfrac{43}{42}\)
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)
=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)
=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)
=> \(5=14x\)
Vậy \(x=\dfrac{5}{14}\)
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)
=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)
=> \(-3x=1\)
Vậy \(x=\dfrac{-1}{3}\)
Mk làm bài 2 thui, bài 1 nhân ra rùi rút gọn đi là đc
a) \(5x^2-5y^2=5\left(x^2-y^2\right)=5\left(x-y\right)\left(x+y\right)\)
b) \(x^2-5xy+x-5y=x\left(x-5y\right)+\left(x-5y\right)=\left(x-5y\right)\left(x+1\right)\)
c) Phần này phải là \(x^2-y^2+4x+4y\)mới đúng, như vậy nó sẽ là :\(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)=\left(x+y\right)\left(x-y+4\right)\)
d) \(x^2-2x-y^2-2y=\left(x^2-y^2\right)-\left(2x+2y\right)=\left(x+y\right)\left(x-y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
Chúc bạn hok tốt !
a) \(x^3-3x^2-4x=0\)
\(\Leftrightarrow x\left(x^2-3x-4\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{0;4;-1\right\}\).
b) \(3x^2-5x-2=0\)
\(\Leftrightarrow3x^2+x-6x-2=0\)
\(\Leftrightarrow x\left(3x+1\right)-2\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};2\right\}\).
a: 3x-15=0
nên 3x=15
hay x=5
b: 4x+20=0
nên 4x=-20
hay x=-5
c: -5x-20=0
nên -5x=20
hay x=-4
\(a.6x-3=5x+2\)
\(\Leftrightarrow6x-3-5-2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(S=\left\{1\right\}\)
\(b.2-3x=5x-6\)
\(\Leftrightarrow2-3x-5x+6=0\)
\(\Leftrightarrow-8x+8=0\)
\(\Leftrightarrow x=1\)
\(S=\left\{1\right\}\)
\(c.\left|3x\right|=2x+7\left(1\right)\)
\(TH_1:3x\ge0\Leftrightarrow x\ge0\)
\(\left(1\right)\Leftrightarrow3x=2x+7\)
\(\Leftrightarrow3x-2x=7\)
\(\Leftrightarrow x=7\left(n\right)\)
\(TH_2:3x< 0\Leftrightarrow x< 0\)
\(\left(1\right)\Leftrightarrow-3x=2x+7\)
\(\Leftrightarrow-3x-2x=7\)
\(\Leftrightarrow-5x=7\)
\(\Leftrightarrow x=\dfrac{-5}{7}\left(n\right)\)
Vậy pt (1) có tập n0 S = \(\left\{7,\dfrac{-5}{7}\right\}\)