Tìm x :
a) x + 15 \(⋮\)x + 8
b) 2x +14 \(⋮\)x + 7
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\(x:\dfrac{13}{16}=\dfrac{5}{-8}\\ x=\dfrac{5}{-8}.\dfrac{13}{16}=\dfrac{65}{-128}\)
\(x.\dfrac{-1}{2}=\dfrac{-4}{5}\\ x=\dfrac{-4}{5}:\dfrac{-1}{2}=\dfrac{8}{5}\)
Câu 14:
a. \(\text{3 + x = - 8}.\) \(\Leftrightarrow x=-8-3=-11.\)
b. \(\text{(35 + x) - 12 = 27}.\)
\(\Leftrightarrow35+x=27+12.\)
\(\Leftrightarrow35+x=39.\Leftrightarrow x=39-35=14.\)
c. \(2^x+15=31.\)
\(\Leftrightarrow2^x=16.\)
\(\Leftrightarrow2^x=2^4.\)
\(\Leftrightarrow x=4.\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
-2x - 40 = (5 - x) - (-15 + 60)
-2x - 40 = (5 - x) - 45
-2x - 40 = -40 - x
-2x - x = 40 - 40
-3x = 0
x = 0
2(x - 4) - 3(x + 7) = 14
2x - 8 - 3x - 21 = 14
-x - 29 = 14
-x = 14 + 29
-x = 43
x = -43
-7(5 - x) - 2(x - 10) = 15
-35 + 7x - 2x + 20 = 15
-15 + 5x = 15
5x = 15 + 15
5x = 30
x = 6
`2x-15 = 17``
`=> 2x = 17 + 15`
`=> 2x = 32`
`=> X = 32 : 2`
`=> x = 16`
`156 - (x + 61) = 82`
`=> x + 61 = 156 - 82`
`=> x + 61 = 74`
`=> x = 13`
`2x - 138 = 2^3 . 3^2`
`=>2x - 138 = 72`
`=> 2x = 210`
`=> x = 105`
bài 2:
`23-3x = 8`
`=> 3x = 23 - 8`
`=> 3x = 15`
`=> x = 5`
`(x-35) - 120 = 0`
`=>(x-35) = 120`
`=> x = 120 +35`
`=> x = 155`
`3^x + 2 = 29`
`=> 3^x = 27`
`=> 3^x = 3^3`
`=> x = 3`
a: Ta có: \(150=5^2\cdot2\cdot3;84=2^2\cdot3\cdot7;30=2\cdot3\cdot5\)
Do đó: ƯCLN(150;84;30)\(=2\cdot3=6\)
150⋮x; 84⋮x; 30⋮x
=>x∈ƯC(150;84;30)
=>x∈Ư(6)
mà x<16
nên x∈{1;-1;2;-2;3;-3;6;-6}
b: \(15=3\cdot5;14=2\cdot7;20=2^2\cdot5\)
Do đó: BCNN(15;14;20)\(=2^2\cdot3\cdot5\cdot7=4\cdot3\cdot5\cdot7=12\cdot35=420\)
x⋮15; x⋮14; x⋮20
=>x∈BC(15;14;20)
=>x∈B(420)
mà 400<=x<=1200
nên x∈{420;840}
c: 9⋮x+1
=>x+1∈{1;-1;3;-3;9;-9}
=>x∈{0;-2;2;-4;8;-10}
d: 2x+7⋮x-2
=>2x-4+11⋮x-2
=>11⋮x-2
=>x-2∈{1;-1;11;-11}
=>x∈{3;1;13;-9}
a) Ta có: x + 15 \(⋮\)x + 8
<=> (x + 8) + 7 \(⋮\)x + 8
Do x + 8 \(⋮\)x + 8 => 7 \(⋮\)x + 8
=> x + 8 \(\in\)Ư(7) = {1; -1; 7; -7}
=> x \(\in\){-7; -9; -1; -15}
Vậy ...
b) Ta có: 2x + 14 = 2(x + 7) \(⋮\)x + 7
=> mọi x thì 2x + 14 \(⋮\)x + 7