Tìm x
a) x-[42+(-28)]=-8
b) 9x-1=9
c) 2x+1.22014=22015
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 89-(73-x) = 20
=> 73+x =89-20
=> 73+x=69
=> x=69-73
=> x=-4
b) (x+7)-25=13
=> x+7=13+25
=> x+7=38
=> x=38-7
=> x=31
c) 98-(x+4)=20
=> x+4=98-20
=> x+4=78
=> x=78-4
=> x=73
d) 140:(x-8)=7
=> x-8=140:7
=> x-8=20
=> x=20+8
=> x=28
e) 4(x+41)=400
=> x+41=400:4
=> x+41=100
=> x=100-41
=> x=59
f) x-[42+(-28)]=-8
=> x-14=-8
=> x=-8+14
=> x=6
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
\(2^{x+1}\cdot2^{2014}=2^{2015}\\ 2^{x+1}=2^{2015}:2^{2014}\\ 2^{x+1}=2\\ =>x+1=1\\ x=1-1\\ x=0\)
a: 3x-2=2x-3
=>x=-1
b: 2x+3=5x+9
=>-3x=6
=>x=-2
c: 5-2x=7
=>2x=-2
=>x=-2
d: 10x+3-5x=4x+12
=>5x+3=4x+12
=>x=9
e: 11x+42-2x=100-9x-22
=>9x+42=78-9x
=>18x=36
=>x=2
f: 2x-(3-5x)=4(x+3)
=>2x-3+5x=4x+12
=>7x-3=4x+12
=>3x=15
=>x=5
\(a,x\left(y-2\right)=8\\ \Rightarrow x;\left(y-2\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(x\) | \(-8\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) | \(8\) |
\(y-2\) | \(-1\) | \(-2\) | \(-4\) | \(-8\) | \(8\) | \(4\) | \(2\) | \(1\) |
\(y\) | \(1\) | \(0\) | \(-2\) | \(-6\) | \(10\) | \(6\) | \(4\) | \(3\) |
Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-4;0\right),\left(-2;-2\right),\left(-1;-6\right),\left(2;6\right),\left(4;4\right),\left(8;3\right)\)
\(b,\left(x-1\right)\left(y-2\right)=9\\ \Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\)
\(x-1\) | \(-9\) | \(-3\) | \(-1\) | \(1\) | \(3\) | \(9\) |
\(y-2\) | \(-1\) | \(-3\) | \(-9\) | \(9\) | \(3\) | \(1\) |
\(x\) | \(-8\) | \(-2\) | \(0\) | \(2\) | \(4\) | \(10\) |
\(y\) | \(1\) | \(-1\) | \(-7\) | \(11\) | \(5\) | \(3\) |
Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-2;-1\right),\left(0;-7\right),\left(2;11\right),\left(4;5\right),\left(10;3\right)\)
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a) √16x = 8 (điều kiện: x ≥ 0)
⇔ 16x = 82 ⇔ 16x = 64 ⇔ x = 4
(Hoặc: √16x = 8 ⇔ √16.√x = 8
⇔ 4√x = 8 ⇔ √x = 2 ⇔ x = 4)
b) điều kiện: x ≥ 0
c) điều kiện: x - 1 ≥ 0 ⇔ x ≥ 1 (*)
x = 50 thỏa mãn điều kiện (*) nên x = 50 là nghiệm của phương trình.
d) Vì (1 - x)2 ≥ 0 ∀x nên phương trình xác định với mọi giá trị của x.
- Khi 1 – x ≥ 0 ⇔ x ≤ 1
Ta có: 2|1 – x| = 6 ⇔ 2(1 – x) = 6 ⇔ 2(1 – x) = 6
⇔ –2x = 4 ⇔ x = –2 (nhận)
- Khi 1 – x < 0 ⇔ x > 1
Ta có: 2|1 – x| = 6 ⇔ 2[– (1 – x)] = 6
⇔ x – 1 = 3 ⇔ x = 4 (nhận)
Vậy phương trình có hai nghiệm: x = - 2; x = 4