Bai 1 tim x
a/(x-3)(x+3)-2x=4x
b/(x-2)(x-3)=x^2-4
c/(x-2)(x+3)+(3+x)(3-x)=7
d/(x-1)(x^2+x+1)-x^3-2x=8
e/(x-2)(x+2)-x(x+4)=5
C)(2x-3)2-(2x-3)(x-10)=7
d)(3x-4)2-9(x-3)(x+3)=8
e)(3x-5)(5-3x)+9(x+1)2=30
f)(x+4)2-(x+1)(x-1)=16
Mình cần gấp
c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)
\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)
\(\Leftrightarrow11x=28\)
hay \(x=\dfrac{28}{11}\)
d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)
\(\Leftrightarrow-24x=-89\)
hay \(x=\dfrac{89}{24}\)
f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
hay \(x=-\dfrac{1}{8}\)
Tìm x
a,x\(^3\)-4x\(^2\)=-4x
b, \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)
\(a,\Leftrightarrow x^3-4x^2+4x=0\\ \Leftrightarrow x\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow4\left(x-1\right)=3x+6\left(2x-3\right)\\ \Leftrightarrow4x-4=3x+12x-18\\ \Leftrightarrow11x=14\Leftrightarrow x=\dfrac{14}{11}\)
a/ \(x^3-4x^2=-4x\)
\(\Leftrightarrow x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b/ \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)
\(\Leftrightarrow8\left(x-1\right)=6x+12\left(2x-3\right)\)
\(\Leftrightarrow8x-8=6x+24x-36\)
\(\Leftrightarrow8x-8=30x-36\)
\(\Leftrightarrow8x-30x=8-36\)
\(\Leftrightarrow-22x=-28\)
\(\Leftrightarrow x=\dfrac{14}{11}\)
Bài 1:Thực hiện phép tính
a,(5-2x)(x+3)-4x(x+2) b,(3x+1)(x-3)-4(x+2)(x-2)
c,3(x-4)(x+3)+(x-5)(x+3) d,2x(x-4)+(3x-1)(2x-5)
Bài 2:Tìm x biết
a,5x(x+3)-(5x+2)(x+3)=7
b,(3x-1)(3x+2)-9(x+2)(x-2)=10
c,(x+1)(2x-5)+2(3-x)(x+2)=7
d,(1-3x)(x+2)+3x(x-5)=8
Tìm x
a,(x+3)^2-(2-x)^2=1
b,5(x-2)^2-(x+3)^2=(2x-1)^2
c,(x-1)^2-x(x+5)^2=7
d,(3x-2)^2-9(x+2)^2=3
a) (x+3)^2 - (2-x)^2 = 1
x^2 + 6x + 9 - (4 - 4x + x^2) = 1
x^2 + 6x + 9 - 4 + 4x - x^2 = 1
10x + 5 = 1
10x = -4
x = -4/10
x = -2/5
Vậy giá trị của x là -2/5.
b) 5(x-2)^2 - (x+3)^2 = (2x-1)^2
5(x^2 - 4x + 4) - (x^2 + 6x + 9) = 4x^2 - 4x + 1
5x^2 - 20x + 20 - x^2 - 6x - 9 = 4x^2 - 4x + 1
4x^2 - 26x + 30 = 4x^2 - 4x + 1
-26x + 30 = -4x + 1
-22x = -29
x = 29/22
Vậy giá trị của x là 29/22.
c) (x-1)^2 - x(x+5)^2 = 7
x^2 - 2x + 1 - x(x^2 + 10x + 25) = 7
x^2 - 2x + 1 - x^3 - 10x^2 - 25x = 7
-x^3 - 9x^2 - 27x - 6 = 0
d) (3x-2)^2 - 9(x+2)^2 = 3
9x^2 - 12x + 4 - 9x^2 - 36x - 36 = 3
-48x - 32 = 3
-48x = 35
x = -35/48
Vậy giá trị của x là -35/48.
bai 1:tim x(chu y dau * la dau nhan)
a)(x+1/4)+(3x-4)+2*(x-3)=1
b)2*(x-3)=3(x+2)-x+1
c)x*(x+3)+x(x-2)=2x*(x-1)
d)(x-1)*3x-2*(x+2)-2x=x(x-1)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
Rút gọn biểu thức:
a) (x+2)(x-2)-(x-3)(x+1)
b) (x2-5)(x+3)+(x+4)(x-x2)
c)(x-5)(2x+3)-2x(x-3)+x+7
d)(2x+1)2+(3x-1)2+2(2x+1)(3x-1)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
Giải phương trình:
a, x^2+3|x|-4=0
b,|x^2-4|=x^2-4
c,(x+1)^2-|3-2x|-|x-2|^2+6=0
d,x^2+4x+3+|2x+5|-(x+1)(x+3) - 5+2x=0
Giải bất phương trình:
a, 2|x-1| <x+1
b, |x-3| > x+1 phần 2
mình đang cần gấp ;-;
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
V)(-1/2x+3)(2x+6-4c^3) F)(2x-5)(x^2-x+3) W)(3x+1)(x^2-2x-5) X)(6x-3)(x^2+x-1) Y)(5x-2)(3x+1-x^2) Z)(3/4x+1)(4x^2+4x+4)
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)
\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)
\(=-x^2-3x+2c^3x+6x+18-12c^3\)
\(=-x^2+3x+2c^3x+18-12c^3\)
f) \(\left(2x-5\right)\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)
\(=2x^3-2x^2+6x-5x^2+5x-15\)
\(=2x^3-7x^2+11x-15\)
w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)
\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)
\(=3x^3-6x^2-15x+x^2-2x-5\)
\(=3x^3-5x^2-17x-5\)
x) \(\left(6x-3\right)\left(x^2+x-1\right)\)
\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)
\(=6x^3+6x^2-6x-3x^2-3x+3\)
\(=6x^3+3x^2-9x+3\)
y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)
\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)
\(=15x^2+5x-5x^3-6x-2+2x^2\)
\(=-5x^3+17x^2-x-2\)
z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)
\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)
\(=3x^3+3x^2+3x+4x^2+4x+4\)
\(=3x^3+7x^2+7x+4\)
f: =2x^3-2x^2+6x-5x^2+5x-15
=2x^3-7x^2+11x-15
w: =3x^3-6x^2-15x+x^2-2x-5
=3x^3-5x^2-17x-5
x: =6x^3+6x^2-6x-3x^2-3x+3
=6x^3+3x^2-9x+3
y: =(5x-2)(-x^2+3x+1)
=-5x^3+15x^2+5x+2x^2-6x-2
=-5x^3+17x^2-x-2
z: =3x^3+3x^2+3x+4x^2+4x+4
=3x^3+7x^2+7x+4
Tìm x
a,(7x+4)^2-(7x+4)(7x-4)=0
b, 5( x + 3 )( x - 3 ) + ( 2x + 3 )^2+(x-6)^=10
c, (x + 1)^3 + (x – 2)^3 – 2x^2 (x – 1,5) = 3
d,( x + 2)(x^2 – 2x + 4)(x – 2)(x^2 + 2x + 4) = – 65
e, 4x^2 + 4x – 5 = 2
f,16x^2 – 9(x + 1)^2 = 0
Các bạn giúp mình vs mai mình phải nộp rùii
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)