6x=-16+2x
2[x-7]=-16
Tìm x
2x-7+(x-14)=0
x^2-6x=0
(x-3)(16-4x)=0
(x-3)-(16-4x)=0
(x-3)+(16-4x)=0
Mấy câu này khá giống nhau nhé anh (câu 1 giống câu 4 và 5, cấu 2 giống câu 3) =)))
Câu 1: 2x - 7 + (x - 14) = 0
<=> 3x -21 = 0
<=> 3x = 21 => x = 7
Câu 2:
x2 - 6x = 0 <=> x.(x - 6) = 0 => \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
Chúc anh học tốt !!!
Câu 1, 2 có người làm rồi nên mik làm tiếp cho mấy câu tiếp. Cứ áp dụng A.B = 0 => A = 0 hoặc B = 0
3; ( x - 3 )( 16 - 4x ) = 0
=> x - 3 = 0 hoặc 16 - 4x = 0
=> x = 3 hoặc x = 4
Vậy x = 3 hoặc x = 4.
4; ( x - 3 ) - ( 16 - 4x ) = 0
=> x - 3 - 16 + 4x = 0
=> ( x + 4x ) - ( 3 + 16 ) = 0
=> 5x - 19 = 0
=> x = 19/5
Vậy x = 19/5
5; ( x + 3 ) + ( 16 - 4x ) = 0
=> x + 3 + 16 - 4x = 0
=> ( x - 4x ) + ( 16 + 3 ) = 0
=> 3x + 19 = 0
=> x = 19/3
Vậy x = 19/3
a) \(\sqrt{2x-5}=2\)
b) \(\sqrt{x^2-6x+9}=7\)
c) \(\sqrt{x^2-8x+16}=4-x\)
\(a,ĐK:x\ge\dfrac{5}{2}\\ PT\Leftrightarrow2x-5=4\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\le4\\ PT\Leftrightarrow\left|x-8\right|=4-x\\ \Leftrightarrow\left[{}\begin{matrix}x-8=4-x\left(x\ge8\right)\\8-x=4-x\left(x\le8\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\left(trái.vs.ĐK\right)\\0x=4\left(ktm\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) \(\sqrt{2x-5}=2\)
\(\Leftrightarrow\) \(\sqrt{2x-5}^2=2^2\)
\(\Leftrightarrow\) \(2x-5=4\)
\(\Leftrightarrow\) 2x = 9
\(\Leftrightarrow\) x = \(\dfrac{9}{2}\)
Chúc bạn học tốt
K.(3x-1)(2x+7)-(x+1)(6x-5)=16
L.4x(x-7)=(x-7)
M.2(x+15)-x^2-5x=0
N.(x-2)^2-16=0
k: =>6x^2+21x-2x-7-6x^2+5x-6x+5=16
=>19x-x-2=16
=>18x=18
=>x=1
l: =>4x(x-7)-(x-7)=0
=>(x-7)(4x-1)=0
=>x=7 hoặc x=1/4
m: =>2x+30-x^2-5x=0
=>-x^2-3x+30=0
=>x^2+3x-30=0
=>\(x=\dfrac{-3\pm\sqrt{129}}{2}\)
n: =>(x-2-2)(x-2+2)=0
=>x(x-4)=0
=>x=0 hoặc x=4
(3x-1)(2x+7)+(x+1)(5-6x)=16
\(\Leftrightarrow6x^2+21x-2x-7+5x-6x^2+5-6x=16\)
\(\Leftrightarrow18x-18=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
(3x-1) .(2x+7)-(x+1) .(6x-5)+16
a, 2x( 3x + 1 ) + 3x ( 4 - 2x ) = 7
b, 4(18 - 5x ) - 12( 3x - 7 ) = 15(2x - 16 ) - 6(x + 14 )
c, ( 3x + 2 ) ( 2x + 9 ) - ( x + 2 ) ( 6x + 1 ) = ( x + 1 ) - ( x - 6 )
Tìm x
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
Tìm x, biết:
a) (2x+2)(x-1)-(x+2)(2x+1)=0;
b)(3x+1)(2x-3)-6x(x+2)=16;
c)(12x-5)(4x-1)+(3x-7)(1-16x)=81
mn ơi giúp mik vs ạ :<
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
Tìm x: a) (3x-1)(2x+7) - (x+1)(6x-5) = 16
b) (2x+3)2 - 2(2x+3)(2x-5) + (2x-5)2 = x2 + 6x + 64
(3x-1)(2x+7)-(x+1)(6x-50)=16. tìm x