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Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a) -16 + 23 + x = -16
=> 7 + x = -16
=> x = -16 - 7
=> x = -23
b) 3x + 17 = 12
=> 3x = 12 - 17
=> 3x = -5
=> x = -5/3
vì x là số nguyên => x ko có gtri thõa mãn đề bài
c) (2x - 5) + 17 = 16
=> 2x - 5 = 16 - 17
=> 2x - 5 = -1
=> 2x = -1 + 5
=> 2x = 4
=> x = 4 : 2
=> x = 2
\(-16+23+x=-16\)
\(\Rightarrow x=-16-23+16\)
\(\Rightarrow x=-23\)
\(3x+17=12\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(24:\left(3x-2\right)=-3\)
\(\Rightarrow3x-2=-8\)
\(3x=-6\)
\(x=-2\)
\(\left(2x-5\right)+17=16\)
\(2x=16-17+5\)
\(x=2\)
\(-6< x< 3\)
\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2\right\}\)
"Tự tính tổng "
\(-\dfrac{3}{5}\times x+\dfrac{1}{2}=\dfrac{4}{5}\)
\(-\dfrac{3}{5}\times x=\dfrac{4}{5}-\dfrac{1}{2}\)
\(-\dfrac{3}{5}\times x=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}:-\dfrac{3}{5}\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(\dfrac{1}{2}\times x+\dfrac{3}{8}=\dfrac{7}{16}\)
\(\dfrac{1}{2}\times x=\dfrac{7}{16}-\dfrac{3}{8}\)
\(\dfrac{1}{2}\times x=\dfrac{1}{16}\)
\(x=\dfrac{1}{8}\)
Vậy \(x=\dfrac{1}{8}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}x=-\dfrac{1}{2}\)
\(x\times\left(\dfrac{2}{3}-\dfrac{5}{6}\right)=-\dfrac{1}{2}\)
\(x\times\dfrac{-1}{6}=-\dfrac{1}{2}\)
\(x=3\)
Vậy \(x=3\)
\(\dfrac{2}{5}x+\dfrac{3}{10}x=-\dfrac{1}{5}\)
\(x\times\left(\dfrac{2}{5}+\dfrac{3}{10}\right)=-\dfrac{1}{5}\)
\(x\times\dfrac{7}{10}=-\dfrac{1}{5}\)
\(x=-\dfrac{2}{7}\)
Vậy \(x=-\dfrac{2}{7}\).
Bài 22:
a: =>-12x+60+21-7x=5
=>-19x+81=5
=>-19x=-76
=>x=4
b: =>30x+60-6x+30-24x=100
=>90=100(loại)
a) \(\dfrac{2x+5}{2x+1}=\dfrac{2x+1+4}{2x+1}=\dfrac{2x+1}{2x+1}+\dfrac{4}{2x+1}=1+\dfrac{4}{2x+1}\)
Để \(\dfrac{2x+5}{2x+1}\in Z\) thì \(\dfrac{4}{2x+1}\in Z\)
\(\Rightarrow4\) ⋮ \(2x+1\)
\(\Rightarrow2x+1\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow2x\in\left\{0;-2;1;-3;3;-5\right\}\)
\(\Rightarrow x\in\left\{0;-1;\dfrac{1}{2};-\dfrac{3}{2};\dfrac{3}{2};-\dfrac{5}{2}\right\}\)
Mà x nguyên \(\Rightarrow\text{x}\in\left\{0;-1\right\}\)
b) \(\dfrac{3x+5}{x+1}=\dfrac{3x+3+2}{x+1}=\dfrac{3\left(x+1\right)+2}{x+1}=\dfrac{3\left(x+1\right)}{x+1}+\dfrac{2}{x+1}=3+\dfrac{2}{x+1}\)
Để \(\dfrac{3x+5}{x+1}\in Z\) thì \(\dfrac{2}{x+1}\in Z\)
\(\Rightarrow2\) ⋮ \(x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3\right\}\)
c) \(\dfrac{3x+8}{x-1}=\dfrac{3x-3+11}{x-1}=\dfrac{3\left(x-1\right)+11}{x-1}=\dfrac{3\left(x-1\right)}{x-1}+\dfrac{11}{x-1}=3+\dfrac{11}{x-1}\)
Để: \(\dfrac{3x+8}{x-1}\in Z\) thì \(\dfrac{11}{x-1}\in Z\)
\(\Rightarrow11\) ⋮ \(x-1\)
\(\Rightarrow x-1\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)
\(\Rightarrow x\in\left\{2;0;12;-10\right\}\)
d) \(\dfrac{5x+12}{x-2}=\dfrac{5x-10+22}{x-2}=\dfrac{5\left(x-2\right)+22}{x-2}=\dfrac{5\left(x-2\right)}{x-2}+\dfrac{22}{x-2}=5+\dfrac{22}{x-2}\)
Để: \(\dfrac{5x+12}{x-2}\in Z\) thì \(\dfrac{22}{x-2}\in Z\)
\(\Rightarrow22\) ⋮ \(x-2\)
\(\Rightarrow x-2\inƯ\left(22\right)=\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
\(\Rightarrow x\in\left\{3;1;4;0;13;-9;24;-20\right\}\)
e) \(\dfrac{7x-12}{x+16}=\dfrac{7x+112-124}{x+16}=\dfrac{7\left(x+16\right)-124}{x+16}=\dfrac{7\left(x+16\right)}{x+16}-\dfrac{124}{x+16}=7-\dfrac{124}{x+16}\)
Để \(\dfrac{7x-12}{x+16}\in Z\) thì \(\dfrac{124}{x+16}\in Z\)
\(\Rightarrow124\) ⋮ \(x+16\)
\(\Rightarrow x+16\inƯ\left(124\right)=\left\{1;-1;2;-2;4;-4;31;-31;62;-62;124;-124\right\}\)
\(\Rightarrow x\in\left\{-15;-17;-14;-18;-12;-20;15;-47;46;-78;108;-140\right\}\)
mk giúp bạn câu cuối nhé:
3|x+2|-5=16
3|x+2|=16+5
3|X+2|=21
|x+2|=21:3
|x+2|=7
=>x+2=7 hoặc x+2=-7
+) với x+2=7 +) với x+2= -7
x=5. x=-9
vậy x€{5,-9}
nếu có TGian mk sẽ giải cho bạn mấy câu trên
cam ơn bạn nhé bạn có giup mình not câu trên trong vong ngay ko
\(\left(3x-5\right)^2=16\)
\(\left(3x-5\right)^2=4^2\) hoặc \(\left(3x-5\right)^2=\left(-4\right)^2\)
\(3x-5=4\) hoặc \(3x-5=-4\)
\(3x=4+5\) hoặc \(3x=-4+5\)
\(3x=9\) hoặc \(3x=1\)
\(x=9:3\) hoặc \(x=1:3\)
\(x=3\) hoặc \(x=\dfrac{1}{3}\)
Vậy \(x=3\) hoặc \(x=\dfrac{1}{3}\)
[3x-5]2=16
[3x-5]2=42
3x-5=4
3x=4+5
3x=9
x=9:3
x=3