help me (2x-3)*(6-2x)=0
Những câu hỏi liên quan
Help me
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
Tìm x:
\(\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\left(\dfrac{-2x+6}{\sqrt{x}-1}\right)=0\)
Help me plsss
ĐKXĐ: x>=0; x<>1
PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)
=>6-2x=0
=>x=3
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Giải phương trình
2x3-x2+3x+6=0
Help me
\(\Leftrightarrow2x^3-3x^2+6x+2x^2-3x+6=0\)
\(\Leftrightarrow x\left(2x^2-3x+6\right)+2x^2-3x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x+6=0\left(vn\right)\end{matrix}\right.\)
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B1: Tìm x, biết
a) ( x2 + 1).(x - 3) > 0
b) (9 - 2x).(2x2 + 7 > 0
c) (x - 3).(x + 7) ≥ 0
d) (6 - x).( x - 2) > 0
e) (3x - 5).(2x - 4) ≤ 0
f) (16 - 2x).(x + 3) < 0
HELP ME !
Tìm x :
a) x^2 - 2x + 3 = 0
b) 2x^2 - x - 3 = 0
Help me !!!
1. Tìm x,biết
a,x^2-2x-3=0
b,2x^2+3=-5x
c,2(x-3)(x+1)=(2x+1)(x-3)-12
d,(2x-1)^2+(2-x)(2x-1)=0
e,-x^2+5x=6
g,(x+3)^2>6x+13
h,(x-3)^2<x^2-5x+1
Please,help me !
Câu a :
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-x+3x-3=0\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)
Câu b :
\(2x^2+3=-5x\)
\(\Leftrightarrow2x^2+3+5x=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\2x+3=0\Rightarrow x=-\dfrac{3}{2}\end{matrix}\right.\)
Mấy câu sau khó quá ko bt làm :)
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x3 +2x - 3 = 0
HELP ME!!!
\(x^3+2x-3=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)
Ta có: \(x^2+x+3=\left[x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+3-\left(\dfrac{1}{2}\right)^2\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\) (1)
Mà \(\left(x-1\right)\left(x^2+x+3\right)=0\) từ (1) \(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy x = 1
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\(x^3-x+3x-3=0\)
\(\Leftrightarrow x\left(x^2-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+3\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vl\right)\end{matrix}\right.\)
vậy \(S=\left\{1\right\}\)
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I : Tìm x
a) ( 2x - 1 ) x -x ( 2x +3 ) =7
b) 3 ( 2x -1 ) - 5 ( x-3 ) + 6 ( 3x - 4 ) = 24
help me
a) (2x - 1) x - x (2x + 3) = 7
<=> x (2x - 1 - 2x - 3) = 7
<=> -4x = 7
<=> x = \(-\dfrac{7}{4}\)
b) 3 (2x - 1) - 5 (x - 3) + 6 (3x - 4) = 24
<=> 6x - 3 - 5x + 15 + 18x - 24 = 24
<=> 19x - 12 = 24
<=> 19x = 36
<=> x = \(\dfrac{36}{19}\)
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Tìm số nguyên x; y biết
x.y = - 21
(x+1).(y+2) = 7
x.(2y+1) = 6
xy - 2x - 2y = 0
HELP ME!!!!!!!!