Tìm x
a)x.(5-2x)-2x.(1-x)=15
b)(3x+2)2+(1+3x).(1-3x)=2
Tìm x:
a)x.(5-2x)-2x.(1-x)=15
b)(3x+2)^2+(1+3x).(1-3x)=2
a)x.(5-2x)-2x.(1-x)=15
x [ 5 - 2x -2.(1-x) ] = 15
x ( 5 - 2x -2 + 2x ) =15
x . 3 =15
x = 5
b)(3x+2)2+(1+3x).(1-3x)=2
9x2+12x+4+1-9x2=2
12x + 5 = 2
12x = -3
x = -1/4
a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)
\(\Leftrightarrow\)\(3x=15\)
\(\Leftrightarrow\)\(x=5\)
b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)
\(\Leftrightarrow\)\(12x+3=0\)
\(\Leftrightarrow\)\(x=-0,25\)
1. Rút gọn biểu thức:
A = (x + 2)2 - (x + 3)(x - 1) + 15
B = (x - 1)(x + 1) - (x + 4)2 - 6
C = (3x - 2)(3x + 2) - (3x - 1)2
D = (2x + 1)2 - (2x - 3)2 + 6x
E = (x - 4)2 - x(x + 2) - 2x + 3
\(A=\left(x+2\right)^2-\left(x+3\right)\left(x-1\right)+15\)
\(A=x^2+4x+4-\left(x^2-x+3x-3\right)+15\)
\(A=\left(x^2-x^2\right)+\left(4x+x-3x\right)+\left(15+3+4\right)\)
\(A=2x+22\)
______________________
\(B=\left(x+1\right)\left(x-1\right)-\left(x+4\right)^2-6\)
\(B=\left(x^2-1\right)-\left(x^2+8x+16\right)-6\)
\(B=\left(x^2-x^2\right)-8x-\left(1+16+6\right)\)
\(B=-8x-23\)
_________________
\(C=\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2\)
\(C=\left[\left(3x\right)^2-2^2\right]-\left(9x^2-6x+1\right)\)
\(C=\left(9x^2-9x^2\right)+6x-\left(4+1\right)\)
\(C=6x-5\)
a) Rút gọn biểu thức A = (x + 2)2 - (x + 3)(x - 1) + 15:
Bắt đầu bằng việc mở ngoặc:
A = (x^2 + 4x + 4) - (x^2 + 2x - 3x - 3) + 15
Tiếp theo, kết hợp các thành phần tương tự:
A = x^2 + 4x + 4 - x^2 - 2x + 3x + 3 + 15
Tiếp tục đơn giản hóa:
A = x^2 - x^2 + 4x - 2x + 3x + 4 + 3 + 15
Kết quả cuối cùng:
A = 5x + 19
b) Rút gọn biểu thức B = (x - 1)(x + 1) - (x + 4)2 - 6:
Bắt đầu bằng việc mở ngoặc:
B = (x^2 - 1) - (x^2 + 4x + 4) - 6
Tiếp theo, kết hợp các thành phần tương tự:
B = x^2 - 1 - x^2 - 4x - 4 - 6
Tiếp tục đơn giản hóa:
B = x^2 - x^2 - 4x - 4 - 6 - 1
Kết quả cuối cùng:
B = -4x - 11
c) Rút gọn biểu thức C = (3x - 2)(3x + 2) - (3x - 1)2:
Bắt đầu bằng việc mở ngoặc:
C = (9x^2 - 4) - (9x^2 - 6x + 1)
Tiếp theo, kết hợp các thành phần tương tự:
C = 9x^2 - 4 - 9x^2 + 6x - 1
Tiếp tục đơn giản hóa:
C = 9x^2 - 9x^2 + 6x - 4 - 1
Kết quả cuối cùng:
C = 6x - 5
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
b) 5(2x-3)+4x(x-2)+2x(3-2x)=0
\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
\(\Leftrightarrow x=\dfrac{15}{8}\)
vậy x=\(\dfrac{15}{8}\)
c)3x(2-x)+2x(x-1)=5x(x+3)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)
\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)
Tìm x
a) -3 1/2 : (4/5-1/2x) = 2^2
b) 2x + 3x = 5
c) -2/3x - 1/3x = -2
d) -2/3 (x+1) - 1/2 = -1/3
a: =>-7/2:(4/5-1/2x)=4
=>4/5-1/2x=-7/2:4=-7/8
=>1/2x=4/5+7/8=67/40
=>x=67/20
b: =>5x=5
=>x=1
c: =>x(-2/3-1/3)=-2
=>-x=-2
=>x=2
d: =>-2/3(x+1)=-1/3+1/2=1/6
=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4
=>x=-1/4-1=-5/4
tìm x
a) (x+2)(x+3)-(x-2)(x+5)=6
b) (3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
tìm x
a,(x+5)(x-5)-x(x+3)=10 b,(2x+3)(2x-3)-4(x+2)^2=5
c,9x(x+5)-(3x+2)(3x-2)=7 d,(x+1)^3-x(x^2+3x-5)=8
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
Tìm x
a) (x + 3)2 + (x + 2)(5 – x) = 1
b/ (2x – 1)2 – ( x – 5)( 4x + 3) = 3
c/ 3x (x – 2) + 4x – 8 = 0
d/ 2x (3x + 5) – 18x – 30 = 0
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Rút gọn biểu thức:
a) (x + 2)(x – 2) – (x + 1)2
b) (2x – 1)(4x2 + 2x + 1) – (2x + 1)( 4x2 – 2x + 1)
3. Tìm x biết:
a) (x + 2)(x2 – 2x + 4) – x(x2 – 2) = 15
b) (x – 1)3 – x(x2 – 3x – 4) = 13
thanks
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
TÌM X
a. 3.(x^2-x+2)-x.(2+3x)=0
b. (x-1)^2 + (x-1)(x+2)=0
c. 2x^3 +3x^2+2x+3=0
d. 2x^2+x=6
\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)