\(\frac{x-1}{3}+\frac{3x-5}{2}+\frac{2x}{9}+\frac{-5x+3}{9}=\frac{210}{420}\)

\(\Leftrightarrow\frac{x-1}{3}+\frac{3x-5}{2}+\frac{-3x+3}{9}=\frac{1}{2}\)

\(\Leftrightarrow\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(-3x+3\right)}{18}=\frac{9}{18}\)

\(\Rightarrow6x-6+27x-45-6x+6=9\)

\(\Leftrightarrow6x+27x-6x=9+6+45-6\)

\(\Leftrightarrow27x=54\)

\(\Rightarrow x=2\)

cái này sai đề bài

\(\dfrac{x-1}{3}\)+\(\dfrac{3x-5}{2}+\dfrac{2x}{9}+\dfrac{-5x+3}{9}=\dfrac{210}{420}\)

\(\Leftrightarrow\)\(\dfrac{x-1}{3}+\dfrac{3x-5}{2}\)\(+\dfrac{-3x+3}{9}\)\(=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{6\left(x-1\right)}{18}+\dfrac{9\left(3x-5\right)}{18}\)\(+\dfrac{2\left(-3x+3\right)}{18}=\dfrac{9}{18}\)

\(\Rightarrow\)\(6x-6+27x-45-6x+6=9\)

\(\Leftrightarrow\)\(6x+27x-6x=9+6+45-6\)

\(\Leftrightarrow27x=54\)

\(\Rightarrow\)\(x=2\)

6: =x^2-7xy+5xy-35y^2

=x(x-7y)+5y(x-7y)

=(x-7y)(x+5y)

7: =x^2-2xy-8xy+16y^2

=x(x-2y)-8y(x-2y)

=(x-2y)(x-8y)

8: =3x^2-6xy-4xy+8y^2

=3x(x-2y)-4y(x-2y)

=(x-2y)(3x-4y)

9: =4x^2+4xy+y^2-16y^2

=(2x+y)^2-16y^2

=(2x+y-4y)(2x+y+4y)

=(2x-3y)*(2x+5y)

10: =2(x^2+5xy+4y^2)

=2(x+y)(x+4y)

11: =5x(x+2y+y^2)

\(\Leftrightarrow\frac{2^{3x^2-3x+1}}{3^{x^2-x+1}}.\frac{3^{2x^2-3x+2}}{5^{2x^2-3x+2}}.\frac{5^{3x^2-4x+3}}{7^{3x^2-4x+3}}.\frac{7^{4x^2-5x+4}}{2^{4x^2-5x+4}}=210^{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{\left(3.5.7\right)^{x^2-x+1}}{2^{x^2-2x+1}}=2^{\left(x-1\right)^2}.\left(3.5.7\right)^{\left(x-1\right)^2}\)

\(\Leftrightarrow105^x=2^{2\left(x-1\right)^2}\)

Lấy Logarit cơ số 2 hai vế, ta được :

\(2\left(x-1\right)^2=\left(\log_2105\right)x\)

\(\Leftrightarrow2x^2-\left(4+\log_2105\right)x+2=0\)

\(\Leftrightarrow x=\frac{\left(2+\log_2105\right)\pm\sqrt{\log^2_2105+8\log_2105}}{4}\)

Vậy phương trình đã cho có 2 nghiệm

\(\left(X+5\right)-\left(X-9\right)=X+2\)\(2\)

\(=>X+5-X+9=X+2\)

\(=>\left(X-X\right)+\left(5+9\right)=X+2\)

 \(=>0+14=X+2\)

\(=>14=X+2\)

\(=>X=12\)

(x+5)-(x-9)=x+2

x+5-x+9=x+2

x-x+14=x+2

12=x(cũng bớt mới về đi 2 đơn vị)

hoặc x=14-2suy ra x=12

làm cau 2 đi mk tích

câu 1:

a) 500-(300)-190+(-210)

= 500-300-190-210

= 200 - 210 -190

=-10 - 190

=-200

b) (-3)³ .5+12.(-6)

= -27.5 -72

=-135 - 72

=-207

c) 15.(-19-4)-19.(15-4)

= 15.(-23) - 19.11

=-345 - 209

=-554

câu 2: tìm x thuộc Z

a) 3x-2=3

=> 3x=3/2

=> x=1/2

b) x chia hết cho 5 và -7<x<11

=> x thuộc {-5;0;5;10}

Câu 1: 

a) Ta có: \(500-\left(300\right)-190+\left(-210\right)\)

\(=500-300-190-210\)

\(=\left(500-300\right)-\left(190+210\right)\)

\(=200-400=-200\)

b) Ta có: \(\left(-3\right)^3\cdot5+12\cdot\left(-6\right)\)

\(=\left(-3\right)^3\cdot5-3\cdot4\cdot3\cdot2\)

\(=-5\cdot3^3-3^2\cdot8\)

\(=3^2\cdot\left(-5\cdot3-8\right)\)

\(=9\cdot\left(-15-8\right)=9\cdot\left(-23\right)=-207\)

c) Ta có: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)

\(=-15\cdot19-15\cdot4-15\cdot19+19\cdot4\)

\(=-30\cdot19+4\cdot4\)

\(=-2\cdot\left(15\cdot19+2\cdot4\right)\)

\(=-2\cdot\left(285+8\right)=-586\)

1) -3-(-x) =5

=> -3 + x = 5

=> x = 5 - (-3)

=> x = 8

2)-9-(x-3)=12

=> x- 3 = (-9) - 12

=> x-3 =-21

=> x = (-21)+3

=> x = -18

3)-10-(x-5)=-5

x-5 = (-10) -(-5)

x-5 = -5

x = -5 +5

x = 0

4)210-(10-x) =110

10-x = 210 -110

10-x = 100

x = 10 -100

x = -90

5)4x-(2x-5)=21

=> 4x-2x + 5 =21

=> x(4-2) + 5 = 21

=> x.2 =21 - 5

=> x.2 = 16

=> x = 16:2

=> x = 8

6)14x-5=8x+10

7)15+5x =3x+30

8)2x-5=15-3x

9)2(3x+5)+3(x+1)=x+220

10)4.(x-2+5(x-3)=18

tên j đấy

1) -3-(-x) =5

=> -3 + x = 5

=> x = 5 - (-3)

=> x = 8

2)-9-(x-3)=12

=> x- 3 = (-9) - 12

=> x-3 =-21

=> x = (-21)+3

=> x = -18

3)-10-(x-5)=-5

x-5 = (-10) -(-5)

x-5 = -5

x = -5 +5

x = 0

4)210-(10-x) =110

10-x = 210 -110

10-x = 100

x = 10 -100

x = -90

5)4x-(2x-5)=21

=> 4x-2x + 5 =21

=> x(4-2) + 5 = 21

=> x.2 =21 - 5

=> x.2 = 16

=> x = 16:2

=> x = 8

6)14x-5=8x+10

7)15+5x =3x+30

8)2x-5=15-3x

9)2(3x+5)+3(x+1)=x+220

10)4.(x-2+5(x-3)=18