(2x+5)-315=120
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a, 315+(125-x)=435
125-x=435-315
125-x=120
x=125-120=5
b, 5(x-12)=120
x-12=120:5
x-12=24
x=24+12=36
c, (3x-7):2=28
3x-7=28.2
3x-7=56
3x=56+7
3x=63
x=63:3=21
a, 215+(125-x)=435
=>125-x=435-315
=>125-x=120
=>x=125-120
=>x=5
b,5x(x-12)=120
=>x-12=120:5
=>x-12=24
=>x=24+12
=>x=36
c,(3x-7):2=28
=>3x-7=28x2
=>3x-7=56
=>3x=56+7
=>3x=63
=>x=63:3
=>x=21
15+3.(120-2x)=315
3(120-2x)=300 ( vì 315-15)
120-2x=300/3=100
2x=120-100=20
x=20/2=10
15+3(120-2x)=315
3(120-2x)=315-15
3(120-2x)=300
120-2x=300:3
120-2x=100
2x=120-100
2x=20
x=20:2
x=10
Chúc p hc tot nke!
Đặt a=x+2
Ta có: \(\left(2a-1\right)\cdot a^2\cdot\left(2a+1\right)=315\)
\(\Leftrightarrow a^2\left(4a^2-1\right)=315\)
\(\Leftrightarrow4a^4-a^2-315=0\)
\(\Leftrightarrow4a^4-12a^3+12a^3-36a^2+35a^2-105a+105a-315=0\)
\(\Leftrightarrow4a^3\left(a-3\right)+12a^2\left(a-3\right)+35a\left(a-3\right)+105\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(4a^3+12a^2+35a+105\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left[4a^2\left(a+3\right)+35\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+3\right)\left(4a^2+35\right)=0\)
mà \(4a^2+35>0\forall x\)
nên \(\left(a+3\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy: S={-5;1}
Pt ban đầu tương đương :
\(\left(4x^2+16x+15\right)\left(x^2+4x+4\right)=315\)
\(\Leftrightarrow\left(4x^2+16x+15\right)\left(4x^2+16x+16\right)=1260\)
Đặt \(t=4x^2+16x+16\left(t\ge0\right)\). Pt đã cho trở thành :
\(\left(t-1\right)t=1260\)
\(\Leftrightarrow\left(t-36\right)\left(t+35\right)=0\)
\(\Leftrightarrow t=36\)
\(\Leftrightarrow4x^2+16x+16=36\)
\(\Leftrightarrow\left(x+2\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
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
Bài làm:
Ta có: \(\left(2x+3\right)\left(x+2\right)^2\left(2x+5\right)=315\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+4\right)^2\left(2x+5\right)=1260\)
Đặt \(2x+4=t\)
\(Pt\Leftrightarrow\left(t-1\right)t^2\left(t+1\right)=1260\)
\(\Leftrightarrow t^4-t^2-1260=0\)
\(\Leftrightarrow\left(t^2+35\right)\left(t^2-36\right)=0\)
Mà \(t^2+35>0\Rightarrow t^2-36=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=6\\t=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x+4=6\\2x+4=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)