(4.3x-4)-2=18
4-(3x-4)-2=18
4 - (3x - 4) - 2 = 18
4 - (3x - 4 ) = 18 + 2
4 - (3x - 4) = 20
3x - 4 = 4 - 20
3x - 4 = -16
3x = -16 + 4
3x = -12
x = -4
Bài 2
a) 4(3x-4)-2=18
\(4\left(3x-4\right)-2=18\\ 4\left(3x-4\right)=20\\ 3x-4=5\\ 3x=9\\ x=3\)
Vậy.........
\(9x^4-4x^2=0\)
\(2x^4-x^2-6=0\)
\(x^4-9x^2+100=0\)
\(x^4-3x^2-54=0\)
\(3x^4-10x^2+3=0\)
\(x^4-7x^2-18=0\)
a: \(\Leftrightarrow x^2\left(9x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(3x-2\right)\left(3x+2\right)=0\)
hay \(x\in\left\{0;\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow2x^4-4x^2+3x^2-6=0\)
\(\Leftrightarrow x^2-2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
d: \(\Leftrightarrow x^4-9x^2+6x^2-54=0\)
\(\Leftrightarrow x^2-9=0\)
=>x=3 hoặc x=-3
tìm x,bt:
a,16-18/x/=-8-14/x/
b,5./x/+7.(2/x/-3)=4.(/x/-4)+25
c,4/3x+2/-16=20
d,8/x-1/-4=12+6/x-1/
e,15-3/4x+2/=3
i,18-4.(6-2/x/)=3.(4/x/+5+29
g,27-5/x-10/=4/x-10/-18
h,5/10+3x/-21-7-2/15+3x/
mn ơi,giúp mk với T^T
Bài 1: Thu gọn :
(x+1).(x+2)-3x.(x-4)
Bài 2: Tìm x:
(3x-4).(x-2)=3x.(x-9)
Bài 3: Chứng minh biểu thức không phụ thuộc vào giá trị của biến:
-3x.(x-4).(x-2)-x^2.(-3x+18)+24x-25
1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)
4(3x-4)-2=18
4(3x-4)-2=18
=> 4(3x-4)=18+2
=> 3x-4=20:4
=> 3x=5+4
=> x=9:3
=> x=3
12x - 16 - 2 = 18
12x = 18 + 16 + 2
12x = 36
x = 36 : 12
x = 3
Chúc bạn học giỏi ^^ :(
4(3x-4)-2=18
4(3x-4) = 18+2
4(3x-4) = 20
(3x-4) = 20 : 4
(3x-4) = 5
3x = 5+4
3x = 9
x = 9 : 3
x = 3
4(3x-4)-2=18
4 ( 3x - 4 ) - 2 = 18
4 ( 3x - 4 ) = 18 + 2
4 ( 3x - 4 ) = 20
( 3x - 4 ) = 20 : 4
( 3x - 4 ) = 5
3x = 5 + 4
3x = 9
x = 9 : 3
x = 3
4(3x-4)-2=18
4(3x-4)=18+2
4(3x-4)=20
3x-4=20:4
3x-4=5
3x=5+4
3x=9
x=9:3
x=3
4(3x.4)=18+2
4(3x-4)=20
3x-4=20:4
3x-4=5
3x=5+4
3x=9
x=9:3
x=3
C=-3x(x-4)(x-2)+x(3x-18)-25
Ta có: \(C=-3x\left(x-4\right)\left(x-2\right)+x\left(3x-18\right)-25\)
\(=-3x\left(x^2-6x+8\right)+3x^2-18x-25\)
\(=-3x^3+18x^2-24x+3x^2-18x-25\)
\(=-3x^3+21x^2-42x-25\)
\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=n\)
\(=\left(n+3\right)n-4\)
\(=n^2+3n-4\)
\(=n^2-n+4n-4\)
\(=n\left(n-1\right)+4\left(n-1\right)\)
\(=\left(n-1\right)\left(n+4\right)\)
\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)
\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)
Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):
\(\left(12n+1\right)n=6\)
\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
tìm x
a) 4(18-5x)-12(3x-7)=15(2x-16)-6(x+14)
b) 5(3x+5)-4(2x-3)=5x+3(2x+12)+1
c) 2(5x-8)-3(4x-5)=4(3x-4)+11
a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)