Tim x biết
(x+ 2/3) . (5/4 - 2x ) > 0
70: 4x + 720 / x = 1/2
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a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
a) \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(\Leftrightarrow\left(4x+28\right).3+55=35.5\)
\(\Leftrightarrow\left(4x+28\right).3+55=175\)
\(\Leftrightarrow\left(4x+28\right).3=175-55\)
\(\Leftrightarrow\left(4x+28\right).3=120\)
\(\Leftrightarrow4x+28=120:3\)
\(\Leftrightarrow4x+28=40\)
\(\Leftrightarrow4x=40-28\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12:4\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(12x-4^3\right).8^3=4.8^4\)
\(\Leftrightarrow12x-4^3=4.8^4:8^3\)
\(\Leftrightarrow12x-4^3=4.8^{4-3}\)
\(\Leftrightarrow12x-4^3=4.8\)
\(\Leftrightarrow12x-4^3=32\)
\(\Leftrightarrow12x-64=32\)
\(\Leftrightarrow12x=32+64\)
\(\Leftrightarrow12x=96\)
\(\Leftrightarrow x=96:12\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
c) \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=8.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x-5\right)=720:40\)
\(\Leftrightarrow41-\left(2x-5\right)=18\)
\(\Leftrightarrow2x-5=41-18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow2x=23+5\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=28:2\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
1) Ta có: 3x - x2 = -(x2 - 3x + 9/4) + 9/4 = -(x - 3/2)2 + 9/4
Ta luôn có: -(x - 3/2)2 \(\le\)0 \(\forall\)x
=> -(x - 3/2)2 + 9/4 \(\le\)9/4 \(\forall\)x
Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2
Vậy Max của 3x - x2 là 9/4 tại x = 3/2
2) Ta có : -(x2 + y2) + x + 3y+ 10 = -x2 - y2 + x + 3y + 10 = -(x2 - x + 1/4) - (y2 -3y + 9/4) + 25/2 = -(x - 1/2)2 - (y - 3/2)2 + 25/2
Ta luôn có: -(x - 1/2)2 \(\le\)0 \(\forall\)x
-(y - 3/2)2 \(\le\)0 \(\forall\)y
=> -(x - 1/2)2 - (y - 3/2)2 + 25/2 \(\le\)25/2 \(\forall\)x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{3}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)
Vậy ...
720 : [41 - (2x - 5)] = 8 . 5
=> 720 : [41 - (2x - 5)] = 40
=> 41 - (2x - 5) = 720 : 40
=> 41 - (2x - 5) = 18
=> 2x - 5 = 41 - 18
=> 2x - 5 = 23
=> 2x = 23 + 5
=> 2x = 28
=> x = 28 : 2
=> x = 14
Vậy x = 14
1 + 2 + 3 + 4 + ... + (x - 1) + x = 78
có x số hạng
=> (x + 1) . x : 2 = 78
=> (x + 1) . x = 78 . 2
=> (x + 1) . x = 156
=> (x + 1) . x = 13 . 12
=> x = 12
Vậy x = 12
a,
128-3x-12=23
3x=128-12-23
3x=93
x=93:3
= 31
b,
(12x+84+55):5=35
12x+84+55=35.5
12x+84+55=175
12x=175-55-84
12x=36
x=36:12
x=3
1) 4x(x-5)-(x-1)(4x-3)=5
<=>4x2-20x-4x2+3x+4x-3=5
<=>-13x=8
<=>x=-8/13
Thôi mỏi tay quá tìm x luôn nha
2) x=1.875
3) x=17/7
\(\left(x+\frac{2}{3}\right)\left(\frac{5}{4}-2x\right)>0\)
th1 :
\(\hept{\begin{cases}x+\frac{2}{3}>0\\\frac{5}{4}-2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{2}{3}\\-2x>-\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x>-\frac{2}{3}\\x>\frac{5}{8}\end{cases}\Rightarrow}x>\frac{5}{8}}\)
th2 :
\(\hept{\begin{cases}x+\frac{2}{3}< 0\\\frac{5}{4}-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{2}{3}\\-2x< -\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x< -\frac{2}{3}\\x< \frac{5}{8}\end{cases}\Rightarrow}x< -\frac{2}{3}}\)
vậy_