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a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
f) (x – 5)2 = 9
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a, 2.x + 7 = 15
2x = 8
x = 4
b, 25 – 3.(6 – x) = 22
3.(6-x) = 3
6-x = 1
x = 5
c, [(2x – 11) : 3 + 1].5 = 20
(2x-11) : 3+1 = 4
(2x-11):3 = 3
2x-11 = 1
2x = 12
x = 6
e, 2 . 3x = 10 . 312 + 8 . 274
6x = 3120 + 2192
6x = 5312
x = 5312/6
g, x – 12 = (–8) + (–17)
x - 12 = -25
x = -13
Lần sau tách nhỏ nội dung câu hỏi ra nha em, chứ trả lời thế này biếng lắm '^^ Chị làm chỉ mang tính tham khảo kết quả thôi, còn cụ thể thì em tách từng bước một ra he :>
a, \(2\cdot x+7=15\)
\(\Leftrightarrow2\cdot x=8\)
\(\Leftrightarrow x=4\)
Vậy x = 4.
b, \(25-3\cdot\left(6-x\right)=22\)
\(\Leftrightarrow3\cdot\left(6-x\right)=3\)
\(\Leftrightarrow6-x=1\)
\(\Leftrightarrow x=5\)
Vậy x = 5.
c, \(\left[\left(2x-11\right):3+1\right]\cdot5=20\)
\(\Leftrightarrow\left(2x-11\right):3+1=4\)
\(\Leftrightarrow\left(2x-11\right):3=3\)
\(\Leftrightarrow2x-11=9\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
Vậy x = 10.
d, \(\left(25-2x\right)\cdot3:5-32=42\)
\(\Leftrightarrow\)\(\frac{3\cdot\left(25-2x\right)}{5}=74\)
\(\Leftrightarrow3\cdot\left(25-2x\right)=370\)
\(\Leftrightarrow25-2x=\frac{370}{3}\)
\(\Leftrightarrow2x=-\frac{295}{3}\)
\(\Leftrightarrow x\approx49\)
Vậy \(x\approx49\) .
e, \(2\cdot3x=10\cdot312+8\cdot274\)
\(\Leftrightarrow6x=5312\)
\(\Leftrightarrow x=5312:6\approx885\)
Vậy \(x\approx885\) .
g, \(x-12=\left(-8\right)+\left(-17\right)\)
\(\Leftrightarrow x-12=-25\)
\(\Leftrightarrow x=-25+12=-13\)
Vậy x = -13.
h, \(7-2x=18-3x\)
\(\Leftrightarrow-2x+3x=18-7\)
\(\Leftrightarrow x=11\)
Vậy \(x=11\) .
i, \(3\cdot\left(x+5\right)-x-11=24\)
\(\Leftrightarrow3x+15-x-11=24\)
\(\Leftrightarrow2x=24+11-15\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\) .
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
1. Thực hiện phép tính bằng cách hợp lí :
a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]
= 0+(-100) = -100
b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75
= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000
c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]
= (-151) + [(-100) + 100] = -151
d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]
= 32+[(-200)+(-400)] = 32+(-600) = -568
e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65
= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)
= 65.(-70) = -4550
g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43
= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)
= 16.(-10) = -160
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
| x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
| y | -3 | 5 | -1 | 3 | 0 | 2 |