Tìm x
a, ( 2x - 3 ) : 3 = 7
b, ( 2x - 23 ) . 2 = 24
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a) \(3\dfrac{4}{5}:\dfrac{8}{5}=0,25:x\)
\(\Rightarrow\dfrac{19}{5}.\dfrac{5}{8}=\dfrac{x}{4}\)
\(\Rightarrow\dfrac{x}{4}=2\Rightarrow x=8\)
b) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Rightarrow x=\dfrac{1}{8}+\dfrac{1}{32}=\dfrac{5}{32}\)
c) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
a)
x - 17 = -5
x = ( - 5) + 17
x = 12
b )
2x + 1/4 = 3/2
2x = 3/2 - 1/4
2x = 6/4 - 1/4
2x = 5/4
x = 5/4 : 2
x = 5/4 x 1/2
x = 5/8
tk và kb với mk nha! mơn ạ!
a)\(x-17=-5\Rightarrow x=-5+17\Rightarrow x=12\)
b)\(2x+\frac{1}{4}=\frac{3}{2}\Rightarrow2x=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)\(\Rightarrow x=\frac{5}{4}:2\Rightarrow x=\frac{5}{8}\)
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
a)
\(13x+3=16\\ 13x=16-3\\ 13x=13\\ x=13:13\\ x=1\)
b)
\(2x-138=24:23\\ 2x-138=\dfrac{24}{23}\\ 2x=\dfrac{24}{23}+138\\ 2x=\dfrac{3198}{23}\\ x=\dfrac{3198}{23}:2\\ x=\dfrac{1599}{23}\)
@ Nguyễn Quốc Đạt em bị nhầm đề bài nhé
2\(x\) - 138 = 24 : 23
2\(x\) - 138 = 2
2\(x\) = 2 + 138
2\(x\) = 140
\(x\) = 140 : 2
\(x\) = 70
A+B+C
=2x^3-5x^2+x-7+x^2-2x+6+C
=2x^3-4x^2-x-1-x^3+4x^2-1
=x^3-x-2
\(-2x-\left(x-17\right)=34-\left(-x+25\right)\)
\(-2x-x+17=34+x-25\)
\(-2x-x-x=34-25-17\)
\(-4x=-8\Leftrightarrow x=2\)
\(17x-\left(16x-37\right)=2x+43\)
\(17x-16x+37=2x+43\)
\(17x-16x-2x=-37+43\)
\(-x=6\Leftrightarrow x=6\)
\(-2x-3\left(x+17\right)=34-2\left(-x+25\right)\)
\(-2x-3x-51=34+2x-50\)
\(-2x-3x-2x=34-50+51\)
\(-7x=35\Leftrightarrow x=-5\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
a. (2x - 3) : 3 = 7
2x - 3 = 7 . 3
2x - 3 = 21
2x = 21 + 3
2x = 24
x = 24 : 2
x = 12
b. (2x - 23) . 2 = 24
2x - 23 = 24 : 2
2x - 23 = 23
2x - 8 = 8
2x = 8 + 8
2x = 16
x = 16 : 2
x = 8
\(a.\left(2x-3\right):3=7\)
\(\left(2x-3\right)=3\cdot7\)
\(\left(2x-3\right)=21\)
\(2x=21+3\)
\(2x=24\)
\(x=24:2\)
\(x=12\)
=>Vậy:\(x=12\)
\(b.\left(2x-2^3\right)\cdot2=2^4\)
\(\left(2x-2^3\right)\cdot2=16\)
\(\left(2x-2^3\right)=16:2\)
\(\left(2x-2^3\right)=8\)
\(2x=8+2^3\)
\(2x=8+8\)
\(2x=16\)
\(x=16:2\)
\(x=8\)
=>Vậy:\(x=8\)