1/- Biết 2x+2 – 2x = 48 . Vậy x =
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(2x2 + 1)(x-3)=0
\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)
Vậy x=3
48-(15-x)5=48
(15-x)5=48-48
(15-x)5=0
=> 15-x =0
x =15-0
x =15
Vậy x=15
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow2x=24\)
\(\Rightarrow x=12\)
(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0
=> 2015.2x + (1 + 2 + 3 + ... + 2015) = 0
=> 4030x + (2015 + 1).2015 : 2 = 0
=> 4030x = -2031120
=> x = -504
(2x+1)+(2x+2)+...........+(2x+2015)=0
2x .2015+(1+2+3+...............2015)=0
4030x + 2031120 =0
4030x =0-2031120
4030x = -2031120
x = -2031120:4030
x = -504
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
a, 36:(x–5) = 2 2
(x–5) = 9
x = 14
b, [3.(70–x)+5]:2 = 46
[3.(70–x)+5] = 92
70–x = 29
x = 41
c, 450:[41–(2x–5)] = 3 2 .5
41–(2x–5) = 10
2x–5 = 31
2x = 36
x = 18
d, 230+[ 2 4 +(x–5)] = 315. 2018 0
16+(x–5) = 315–230
x–5 = 85–16
x = 69+5
x = 74
e, 2 x + 2 x + 1 = 48
2 x .(2+1) = 48
2 x = 16 = 2 4
x = 4
f, 3 x + 2 + 3 x = 2430
3 x . 3 2 + 1 = 2430
3 x = 2430:10 = 243 = 3 5
x = 5
a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
a) \(\left(2x-1\right)^2-2x+1=0\)
\(\Leftrightarrow4x^2-4x+1-2x+1=0\)
\(\Leftrightarrow4x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4.4.2=4\)
\(x_1=\dfrac{1}{2};x_2=1\)
b) \(\left(4x-3\right)^2-12x+9=0\)
\(\Leftrightarrow16x^2-24x+9-12x+9=0\)
\(\Leftrightarrow16x^2-36x+18=0\)
\(\Delta=\left(-36\right)^2-4.16.18=144\)
\(x_1=\dfrac{3}{4};x_2=\dfrac{3}{2}\)
c) \(3x^2-48=0\)
\(\Leftrightarrow3x^2=48\)
\(\Leftrightarrow x^2=16\)
\(x_1=4;x_2=-4\)
Tick cho mk nha
=>(2x-1)^2=24^2
=>2x-1=24 hoặc 2x-1=-24
=>x=-23/2 hoặc x=25/2