Tìm x
a, -2/3 . x + 4 = 12
b, -3/4 + 1/4 : x = -3
c, ( 2x - 1/3 ) ( 0.5x +0,25 ) = 0
tìm x biết
a)2 ( x - y )= 12
b)5x (x - 3 )-2x +6=0
c)7x (x - 4)-2 (x - 4)=0
d)(x -1 )(x +1 ) -x(x +3)=
Lời giải:
a. Đề có cả x,y. Bạn xem lại
b.
PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$
$\Leftrightarrow (x-3)(5x-2)=0$
$\Leftrightarrow x-3=0$ hoặc $5x-2=0$
$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$
c.
PT $\Leftrightarrow (7x-2)(x-4)=0$
$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$
$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$
d. Đề thiếu.
x+2=-5
x2+2x-3=0
x+4=10
-3x+2=-7
0.5x+4=-1
x+2=-5
x =-7
x2+2x-3=0
(=) x2-x+3x-3=0
(=) x(x-1)+3(x-1)=0
(=) (x-1)(x+3)=0
=> x-1 = 0 hoặc x+3 = 0
x = 1 x =-3
x+4 = 10
x = 6
-3x + 2 =-7
(=) -3x =-9
(=) x = 3
0.5x + 4 = -1
(=) 0.5x =-5
(=) x =-10
Chúc bạn học tốt ^^
a)x+2=-5
\(\Leftrightarrow\)x =-7
b)x2+2x-3=0
\(\Leftrightarrow\)x2-x+3x-3=0
\(\Leftrightarrow\)x(x-1)+3(x-1)=0
\(\Leftrightarrow\)(x+3)(x-1)=0
\(\Rightarrow\)x+3=0 hoặc x-1=0
\(\Rightarrow\)x=-3 hoặc x=1
c)x+4=10
\(\Leftrightarrow\)x=6
d)-3x+2=-7
\(\Leftrightarrow\)-3x =-7-2
\(\Leftrightarrow\)3x =-9
\(\Leftrightarrow\)x =-3
e)0,5x+4=-1
\(\Leftrightarrow\)0,5x =-1-4
\(\Leftrightarrow\)0,5x = -5
\(\Leftrightarrow\)x =-5:0,5
\(\Leftrightarrow\)x =-10
Tìm x
a. 4(x-3)^2-(2x-1)(2x+1)=10
b. x^3-25x=0
\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+27=10\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{27}{24}\)
Vậy \(x=\dfrac{27}{24}\)
\(b,x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm5\right\}\)
Bài 4: Tìm x
a, (x + 1)2 - x (x - 3) = 2x + 3
b, (x + 2) (x - 3 ) - (x - 3)2 = 15
c, (x + 1)3 -(x3 + 3x2 + 2x - 3) = 0
\(a,\Leftrightarrow x^2+2x+1-x^2+3x-2x=3\\ \Leftrightarrow3x=2\Leftrightarrow x=\dfrac{3}{2}\\ b,\Leftrightarrow x^2-x-6-x^2+6x-9=15\\ \Leftrightarrow5x=30\Leftrightarrow x=6\\ c,\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=0\\ \Leftrightarrow x=-4\)
a) \(\left(x+1\right)^2-x\left(x-3\right)=2x+3\Rightarrow x^2+2x+1-x^2+3x=2x+3\)
\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
b) \(\left(x+2\right)\left(x-3\right)-\left(x-3\right)^2=15\Rightarrow x^2-3x+2x-6-\left(x^2-6x+9\right)=15\)
\(\Rightarrow5x=30\Rightarrow x=6\)
Câu 1:): Tìm X
A) X + 2/3 = 4/5
B) 7/9 - X = 1/3
C) X : 2/3 = 9/8
D) X x 4/7 - X x 3/4 = 10
GIẢI GIÚP MÌNH VỚI Ạ!
A) \(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\)
\(x=\dfrac{22}{15}\)
b)\(\dfrac{7}{9}-x=\dfrac{1}{3}\\ x=\dfrac{7}{9}-\dfrac{1}{3}\\ x=\dfrac{4}{9}\)
C)\(x:\dfrac{2}{3}=\dfrac{9}{8}\\ x=\dfrac{9}{8}x\dfrac{2}{3}\\ x=\dfrac{3}{4}\)
bài 1 tìm x
a)6x^2-72x=0
b)-2x^4+16x=0
c)x(x-5)-(x-3)^2=0
d)(x-2)^3-(x-2)(x^2+2x+4)=0
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
2. Tìm x
a. \(\dfrac{4}{5}-3.\left|x\right|=\dfrac{1}{5}\) b. \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
c. (2x-8)(10-5x)=0 d. \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
Tìm x
a) 4(x + 1)2 + (2x + 1)2 - 8(x – 1)(x + 1) - 11=0
b)(x + 3)2 – (x – 4)(x + 8) – 1 = 0
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20
bài 2: tìm x
a)x-7=12
b)9+4.(x-5)=13
c)(x+2)3=64
a/ x - 7 = 12
=> x = 12 + 7 = 19
b/ 9 + 4(x - 5) = 13
=> 4x - 20 = 4
=> 4x = 24
=> x = 6
c/ (x+2)3 = 64
=> x + 2 = 4
=> x = 2
a) x-7=12
x=12+7
x=19
b) 9+4.(x-5)=13
4.(x-5)=13-9
4.(x-5)=4
(x-5)=4:4
(x-5)=1
x=5+1
x=6
cau cuoi mik ko bt lam nha!
chuc ban hoc tot
a/ x - 7 = 12
=> x = 12 + 7 = 19
b/ 9 + 4(x - 5) = 13
=> 4x - 20 = 4
=> 4x = 24
=> x = 6
c/ (x+2)3 = 64
=> x + 2 = 4
=> x = 2