Tìm \(x\), biết :
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
Tìm x biết:
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
Mình đăng giùm chị mình thôi!
( 12x-5)( 4x-1)+ ( 3x-7)(1-16x)
=> 48x^2-48x^2-12x-20x+3x+112x+5-7=81
=> 83x-2=81
=> x=1
(12x - 5)(4x - 1) + (3x - 7)(1 - 16x) = 81
48x2 - 12x - 20x + 5 + 3x - 48x2 - 7 + 112x = 81
83x - 2 = 81
83x = 81 + 2
83x = 83
x = 1
Tìm x, biết:
a) \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)
b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
Giải:
a) \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1+2x+3\right)\left(3x-1-2x-3\right)=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=4\end{matrix}\right.\)
Vậy ...
b) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-20x-12x+5+3x-7-48x^2+112x=81\)
\(\Leftrightarrow83x-2=81\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy ...
a) (12x-5)(4x-1)+(3x-7)(1-16x)=81
b) (2x-3)(2x+3)-(4x+1).x=1
c) \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
Tìm x
\(\left(4x-1\right)^3+\left(3-4x\right)\left(9+12x+16x^2\right)=\left(8x-1\right)\left(8x+1\right)-\left(3x-5\right)\)
GIÚP MIK VS
Tìm x, biết:
\(\left(12x-5\right).\left(4x-1\right)+\left(3x-7\right).\left(1-16x\right)=81\)
@Liana, @Love Seven giúp mik!!!!!
\(\left(12x-5\right).\left(4x-1\right)+\left(3x-7\right).\left(1-16x\right)=81\)
\(48x^2-12x-20x+5+3x-48x^2-7=112x=81\)
\(83x-2=81\)
\(83x=83\)
=>\(x=1\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
Giải phương trình
a) \(\left(x+2\right)^2-2\left(x-3\right)=\left(x+1\right)^2\)
b)\(3x\left(5-2x\right)-4\left(x+2\right)=6\left(x-2\right)^2\)
c)\(\left(12x-5\right)\left(4x-1\right)-\left(3x-7\right)\left(1+16x\right)=0\)
d)\(\left(2x+7\right)^2=\left(x+3\right)^2\)
bài dễ cậu tự làm được mÀ
Tìm x biết
\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-700=0\)
pt tương đương:
[(4x+1)(3x+2)][(12x-1)(x+1)]=700
<=>(12x2+11x+2)(12x2+11x-1)=700
Đặt t=12x2+11x pt trở thành
(t+2)(t-1)=700
<=>t2+t-702=0
<=>t=26 hoặc t=-27
với t=26 =>x=13/12 hoặc x=-2
với t=-27 không có x
tìm x, biết:
\(a,-4x\left(x-7\right)+4x\left(x^2-5\right)\)\(=28x^2-13\)
\(b,\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)\))\(=\left(-4+x\right)\left(-2x+3\right)+12x^2+2x^2\)
a) -4x(x - 7) + 4x(x2 - 5) = 28x2 - 13
=> -4x2 + 28x + 4x2 - 20x = 28x2 - 13
=> (-4x2 + 4x2) + (28x - 20x) = 28x2 - 13
=> 8x = 28x2 - 13
=> 8x - 28x2 + 13 = 0
=> phương trình vô nghiệm
b) (4x2 - 5x)(3x + 2) - 7x(x + 5) = (-4 + x)(-2x - 3) + 12x2 + 2x2
=> 4x2(3x + 2) - 5x(3x + 2) - 7x2 - 35x = -4(-2x - 3) + x(-2x - 3) + 14x2
=> 12x3 + 8x2 - 15x2 - 10x - 7x2 - 35x = 8x + 12 - 2x2 - 3x + 14x2
=> 12x3 + (8x2 - 15x2 - 7x2) + (-10x - 35x) = (8x - 3x) + 12 + (-2x2 + 14x2)
=> 12x3 - 14x2 - 45x = 5x + 12 + 12x2
=> 12x3 - 14x2 - 45x - 5x - 12 - 12x2 = 0
=> 12x3 + (-14x2 - 12x2) + (-45x - 5x) - 12 = 0
=> 12x3 - 26x2 - 50x - 12 = 0
Làm nốt
Cái câu b sửa cái đề lại nhé dấu " = " ở chỗ (-2x = 3) là gì vậy?
a, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3+13=0\)( vô nghiệm )