rut gon cac bieu thuc
a) \(4x^2\left(5x^2-2x+3\right)\)
b)\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
c)\(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
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
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
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
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
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`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`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`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Tính:
\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)
\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)
\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)
\(d)\left(9x^2-4\right):\left(3x+2\right)\)
\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
Giải các phương trình sau :
a) \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
b) \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
c) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
d) \(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)
e) \(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
f) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
* Tìm x :
a, \(\left(3x-2\right)^2-\left(3x-5\right).\left(3x+2\right)=11\)
b, \(\left(4x-3\right)^2-\left(4x-5\right).\left(4x+5\right)=32\)
c, \(\left(5x-2\right)^2-\left(5x+3\right).\left(5x-5\right)=1\)
d, \(\left(x-4\right)^2-\left(x-7\right).\left(2x-3\right)=5-x^2\)
a) \(\left(3x-2\right)^2-\left(3x-5\right)\left(3x+2\right)=11\)
\(\Leftrightarrow\left(9x^2-12x+4\right)-\left(9x^2+6x-15x-10\right)=11\)
\(\Leftrightarrow9x^2-12x+4-9x^2-6x+15x+10=11\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
b) \(\left(4x-3\right)^2-\left(4x-5\right)\left(4x+5\right)=32\)
\(\Leftrightarrow\left(16x^2-24x+9\right)-\left(16x^2-25\right)=32\)
\(\Leftrightarrow16x^2-24x+9-16x^2+25=32\)
\(\Leftrightarrow-24x+2=0\)
\(\Leftrightarrow-24x=-2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy \(S=\left\{\dfrac{1}{12}\right\}\)
c) \(\left(5x-2\right)^2-\left(5x+3\right)\left(5x-5\right)=1\)
\(\Leftrightarrow\left(25x^2-20x+4\right)-\left(25x^2-25x+15x-15\right)=1\)
\(\Leftrightarrow25x^2-20x+4-25x^2+25x-15x+15=1\)
\(\Leftrightarrow-10x+18=0\)
\(\Leftrightarrow-10x=-18\)
\(\Leftrightarrow x=\dfrac{9}{5}\)
Vậy \(S=\left\{\dfrac{9}{5}\right\}\)
d) \(\left(x-4\right)^2-\left(x-7\right)\left(2x-3\right)=5-x^2\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(2x^2-3x-14x+21\right)=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21-5+x^2=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy \(S=\left\{\dfrac{10}{9}\right\}\)
thực hiện phép tính
a.\(5x^2-3x\left(x+2\right)\)
b.\(3x\left(x-5\right)-5x\left(x+7\right)\)
c.\(3x^2y.\left(2x^2-y\right)-2x^2.\left(2x^2y-y^2\right)\)
d.\(3x^2.\left(2y-1\right)-\left[2x^2.\left(5y-3\right)-2x.\left(x-1\right)\right]\)
e.\(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
f.\(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
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
Tìm x biết : (đề không sai)
1.\(-4x\left(x-7\right)+4x\left(x^2-5\right)\) \(=28x^2-13\)
2.\(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x-7\right)\)= \(\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
3.\(\left(-4x^2-3\right)\left(2x+5\right)-\left(8x-3\right)\) \(\left(-x^2+2\right)=-5x^2\left(x-6\right)-3x^2-4\)
4.\(\left(x-7\right)\left(x+5\right)-\left(x-3\right)\left(x-2\right)\) \(=15x^2\left(x+1\right)-\left(3x^2-1\right)\) \(\left(5x^2-2\right)-21x^2\)
5.\(\left(x-3\right)\left(-x+10\right)+\left(x-8\right)\left(x+3\right)\) \(=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)
6.\(\left(-2x^2+5\right)\left(-x+3\right)-x^2\left(2x-6\right)\) \(=\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+4\right)\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm