Tìm \(x\) :
\(\left(x+3\right)-\left(2x-1\right)=5\)
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1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a) \(x-2=\left(x-2\right)^2\)
\(\left(x-2\right)^2-\left(x-2\right)=0\)
\(\left(x-2\right)\left(x-2-1\right)=0\)
\(\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x-3=0\)
*) \(x-2=0\)
\(x=2\)
*) \(x-3=0\)
\(x=3\)
Vậy \(x=2;x=3\)
b) \(x+5=2\left(x+5\right)^2\)
\(2\left(x+5\right)^2-\left(x+5\right)=0\)
\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)
\(\left(x+5\right)\left(2x+10-1\right)=0\)
\(\left(x+5\right)\left(2x+9\right)=0\)
\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)
*) \(x+5=0\)
\(x=-5\)
*) \(2x+9=0\)
\(2x=-9\)
\(x=-\dfrac{9}{2}\)
Vậy \(x=-5;x=-\dfrac{9}{2}\)
c) \(\left(x^2+1\right)\left(2x-1\right)+2x=1\)
\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)
\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x^2+1+1\right)=0\)
\(\left(2x-1\right)\left(x^2+2\right)=0\)
\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)
*) \(2x-1=0\)
\(2x=1\)
\(x=\dfrac{1}{2}\)
*) \(x^2+2=0\)
\(x^2=-2\) (vô lí)
Vậy \(x=\dfrac{1}{2}\)
d) Sửa đề:
\(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2+3+1\right)=0\)
\(\left(x+1\right)\left(x^2+4\right)=0\)
\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)
*) \(x+1=0\)
\(x=-1\)
*) \(x^2+4=0\)
\(x^2=-4\) (vô lí)
Vậy \(x=-1\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
(x-3)(-2x+5)-2x(x-4)+(x-3)=(x-2)(x-1)-(x2-5x)
<=>-2x2+11x-15-2x2+8x+x-3=x2-3x+2-x2+5x
<=>-4x2+20x-18=2x+2
<=>-4x2+20x-18-2x-2=0
<=>-4x2+18x-20=0
<=>-4x2+8x+10x-20=0
<=>-4x.(x-2)+10.(x-2)=0
<=>(x-2)(-4x+10)=0
<=>x-2=0 hoặc -4x+10=0
<=>x=2 hoặc x=5/2
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
a) \(\left(2x-1\right)\left(2x+1\right)-4x^2=3\Leftrightarrow\left(4x^2-1\right)-4x^2=3\Rightarrow-1=3\) (không đúng)
Tí làm tiếp nhé ;) h đi chơi đã
\(\left(x+3\right)-\left(2x-1\right)=5\\ \Rightarrow x+3-2x+1=5\\ \Rightarrow4-x=5\\\Rightarrow x=4-5\\ \Rightarrow x=-1 \)
Vậy \(x=-1\)
(x + 3) - (2x - 1) = 5
(x + 3) - 2x + 1 = 5
-x + 4 = 5
-x = 5 - 4
-x = 1
-x = -(-1)
⇒ x = - 1
Vậy x = -1