x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

2: \(\Leftrightarrow\left(\dfrac{x+1}{12}-1\right)+\left(\dfrac{x+2}{13}-1\right)=\left(\dfrac{x+3}{14}-1\right)+\left(\dfrac{x+4}{15}-1\right)\)

=>x-11=0

=>x=11

3: \(\Leftrightarrow\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

=>x-100=0

=>x=100

dài quá à :(

1) x - 43 = (35 - x) - 48

=> x + x = 35 - 48 + 43

=> x + x = 30

=> x = 30 : 2

=> x = 15

2) 305 - x + 14 = 48 + (x + 23)

=> 305 - x + 14 = 48 + x + 23

=> -x - x = 48 + 23 - 14 - 305

=> -x - x = -248

=> -x = -248 : 2

=> -x = -124

=> x = 124

3) - (x - 6 + 85) = (x + 51) - 54

=> -x + 6 - 85 = x + 51 - 54

=> -x - x = 51 - 54 + 85 - 6

=> -x - x = 76

=> -x = 76 : 2

=> -x = 38

=> x = -38

4) - (35 - x - 37 - x) = 33 - x

=> -35 + x + 37 + x = 33 - x

=> x + x + x = 33 + 35 - 37

=> x + x + x = 31

=> x = 31 : 3

=> x \(=\dfrac{31}{3}\)

Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.

5) 13 - | x | = | -4 |

=> 13 - |x| = 4

=> |x| = 13 - 4

=> |x| = 9

=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

6) | x | - 3 + 6 = 16

=> |x| = 16 - 6 + 3

=> |x| = 13

=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)

7) 35 - | 2x - 1 | = 14

=> |2x - 1| = 35 - 14

=> |2x - 1| = 21

=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)

8) | 3x - 2 | + 5 = 9 - x

=> |3x - 2| = 9 - 5 - x

=> |3x - 2| = 4 - x

=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)

Vì x \(\in\) Z nên x = -1.

9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )

=> x - (-18) > 12 - 1

=> x + 18 > 11

=> x > 11 - 18

=> x > -7

10) | 17 + ( x - 15 ) | < 4

=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)

11) x2 - 5x = 0

=> x . (2 - 5) = 0

=> x . (-3) = 0

=> x = 0 : (-3)

=> x = 0

12) | x-9 | . (-8) = -16

=> |x - 9| = (-16) : (-8)

=> |x - 9| = 3

=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)

13) | 4 - 5x | = 24 với x < hoặc = 0

=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vì x \(\le\) 0 nên x = -4

14) x . ( x - 2 ) > 0

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x>2\\x< 2\end{matrix}\right.\)

15) x . ( x - 2 ) < 0

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)

16) (x-1) . (y+1) = 5

=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)

17) x . ( y +2 ) = -8

=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=2\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-2\\y+2=4\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=4\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-4\\y+2=2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=2-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

18) xy - 2x - 2y = 0

=> x . (y - 2) - 2y = 0

=> x . (y - 2) - 2y - 4 = -4

=> x . (y - 2) - 2 . (y - 2) = -4

=> (y - 2) . (x - 2) = -4

=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=2\\x-2=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2+2\\x=-2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=4\\x=0\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=-2\\x-2=2\end{matrix}\right.=>\left[{}\begin{matrix}y=-2+2\\x=2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)

19) 2x - 5 \(⋮\) x - 1

=> (2x - 2) - (5 - 2) \(⋮\) x - 1

=> 2(x - 1) - 3 \(⋮\) x - 1

Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1

=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}

=> x \(\in\) {-2; 0; 2; 4}

P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~

A = ( 15/22 - 2/22 ) : 1/33 - ( 6/84 - 8/84) : 1/35 + 1 + 1/5 . ( 3/12 - 2/12 - 12/12 ) . 5/11

A = 13/22 . 33 - (-1/42) . 35 + 1 + 1/5 . - 11/12 . 5/11

A = 39/2 - ( -5/6 ) + 1 + - 11/60 .5/11

A = 39/2 + 5/6 + 1 + (- 1/12)

A = 234/12 + 10 /12 + 12/12 + (-1/12)

A = 255/12

b.

ĐKXĐ: \(x\ge-1\)

\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)

\(\Leftrightarrow...\)

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+2a^2=-b^2+b+3ab\)

\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)

\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)

\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

\(\Leftrightarrow x\sqrt{2x+3}-\sqrt{2x+3}+3-3x+3\sqrt{x+5}-\sqrt{\left(2x+3\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\sqrt{2x+3}\left(x-1\right)-3\left(x-1\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\sqrt{2x+3}-3\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left(x-1-\sqrt{x+5}\right)\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{x+5}=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5-\sqrt{x+5}-6=0\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-2\left(loại\right)\\\sqrt{x+5}=3\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow...\)

cái này tính cái gì thế

ko hiểu

a: =>x/27+1=-2/3

=>x/27=-5/3

=>x=-45

b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)

=>x=221/50

c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)

=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)

=>x=-37/24

e: =>-3/7x=84/45

=>x=-196/45

f: =>11/10x=-2/3

=>x=-20/33

Bài 1:tính 

a)65.(-19)+19.(-35)

=65.(-19)+(-19).35

=(-19).(65+35)

=(-19).100

=-1900

b)85.(35-27)-35.(85-27)

=85.35-85.27-35.85+35.27

=(85.35-35.85)+(-85.27+35.27)

=27.(-85+35)

=27.(-50)

=1350

c)47.(45-15)-47.(45+15)

=47.[(45-15)-(45+15)]

=47.[30-60]

=47.(-30)

=-1410

Bài2: Tìm các số nguyên x biết

a)(-2).(x+6)+6.(x-10)=8

-2x-12+6x-60=8

4x-72=8

4x=72+8

4x=50

 x=\(\frac{25}{2}\)

b)(-4).(2x+9)-(-8x+3)-(x+13)=0

-6x-36+8x-3-x-13=0

x-41=0

     x=41

Bài 1: Tính

a) \(65.\left(-19\right)+19.\left(-35\right)\)

= \(-1235+-665\)

= \(-1900\)

b) \(85.\left(35-27\right)-35.\left(85-27\right)\)

= \(-1350\)

c) \(47.\left(45-15\right)-47.\left(45+15\right)\)

=\(-1410\)

Bài 2: Tìm x:

\(\left(-2\right).\left(x+6\right)+6.\left(x-10\right)=8\)

\(x=20\)

Cảm ơn các bạn nha

\(\left|x\right|-3+6=16\\ \left|x\right|=13\Rightarrow x=\pm13\)

\(35-\left|2x-1\right|=14\\ \left|2x-1\right|=21\\ \left(2x-1\right)^2=21^2\\ \left(2x-1+21\right)\left(2x-1-21\right)=0\\ 4\left(x+10\right)\left(x-11\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-10\\x=11\end{matrix}\right.\)

\(\left|3x-2\right|+5=9-x\\ \left|3x-2\right|=4-x\\ \Rightarrow\left(3x-2\right)^2=\left(4-x\right)^2\\ \left(3x-2-4+x\right)\left(3x-2+4-x\right)=0\\ 4\left(2x-3\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)