2/3x-50%+x=1/10
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
a, ( x2 - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 4x2 - 4x2 - 16x + 16x + 64 - x ( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
Vậy x = 32 là nghiệm của pt.
b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50
<=> 8x - 24x2 + 2 - 6x + 24x2 - 60x - 4x + 10 = - 50
<=> - 62x + 12 = - 50
<=> - 62x = - 62
<=> x = 1
Vậy x = 1 là nghiệm của pt.
![](https://rs.olm.vn/images/avt/0.png?1311)
e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)
\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)
\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)
\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( x2 - 4x + 16 )( x + 4 ) - x( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 43 - x( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = -50
<=> 2x - 24x2 + 2 + 24x2 - 64x + 10 = -50
<=> -62x + 12 = -50
<=> -62x = -62
<=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)
\(\Leftrightarrow9x+3=11\)
\(\Leftrightarrow9x=11-3\)
\(\Leftrightarrow9x=8\)
\(\Leftrightarrow x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)
\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-50-12\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=\dfrac{-62}{-62}\)
\(\Leftrightarrow x=1\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(x^2+x+3x+3-x^2+5x=11\)
\(x+8x+3=11\)
\(x+8x=8\)
\(x\left(8+1\right)=8\)
\(x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(-62x+12=-50\)
\(-62x=-62\)
\(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)
\(\Leftrightarrow x^2+3x-12-10x-30=0\)
\(\Leftrightarrow x^2-7x-42=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};1\right\}\)
LÀM LUN NHA
2/3x-1/2+x=1/10
5/3x=1/10+1/2
5/3x=3/5
X=9/25
**** CHO MK NHA