\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

a)

\(\frac{20}{11}.\frac{33}{23}.\frac{69}{180}\)

\(=\left(\frac{20}{11}.\frac{69}{180}\right).\frac{33}{23}\)

\(=\frac{23}{33}.\frac{33}{23}=1\)

b)  0,8 x 96 + 0,8 x 2 x 2

=0,8 x ( 96+2+2)

=0,8 x 100

= 80

a) 20/11 x 33/23 x 69/180 = 60/23 x 69/180 = 1

b) 0,8 x 96 + 0,8 x 2 x 2 = 0,8 x ( 96 + 2 ) x 2 = 0,8 x 98 x 2 = 78,4 x 2 = 156,8

a)=\(\frac{20}{11}\).\(\frac{33}{23}\).\(\frac{23}{60}\)

=\(\frac{20}{11}\).(\(\frac{33}{23}\).\(\frac{23}{60}\))

=\(\frac{20}{11}\).\(\frac{759}{1380}\)

=\(\frac{20}{11}\).\(\frac{11}{20}\)

=1

b)=0,8.96+0,8.4

=0,8.(96+4)

=0,8.100

80

`A)2/3=x/60`

`=>40/60=x/60`

`=>x=40`

`B)-1/2=y/18`

`=>-9/18=y/18`

`=>y=-9`

`C)3/x=y/35=-36/84`

Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`

`=>3/x=-3/7`

`=>x=-7`

`y/35=-3/7=-15/35`

`=>y=-15`

`D)7/x=y/27=-42/54`

Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`

`=>7/x=-7/9`

`=>x=-9`

`y/27=-7/9=-21/27`

`=>y=-21`

a) Ta có: \(\dfrac{2}{3}=\dfrac{x}{60}\)

\(\Leftrightarrow x=\dfrac{2\cdot60}{3}=40\)

Vậy: x=40

A/

12= 2².3

60=2².3.5

ƯCLN(12,60)=2².3.5=60

B/

24=2³.3

88=2³.11

ƯCLN(24,88)=2³.3.11=264

C/

96=2⁵.3

224= 2⁵.7

ƯCLN(96,224)=2⁵.3.7=672

D/

34=2.17

96=2⁵.3

ƯCLN(34,96)=2⁵.3.17=1632

Chúc bạn học tốt

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hỏi ngu thế

 a) 3x2 – 7x + 2

\(=3x^2-6x-x+2\)

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

\(=a\left(x^2+1\right)-x\left(a^2+1\right)\)

.......?

a) Ta có: \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

\(=3x\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

\(=x^2a+a-a^2x-x\)

\(=\left(x^2a-a^2x\right)+\left(a-x\right)\)

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

\(=\left(x^2+7x\right)\left(x^2+7x+16\right)+6\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

Tìm x:

a, 96 - 3 . ( x + 1 ) = 42

           3 . ( x + 1 ) = 96 - 42

           3 . ( x + 1 ) = 54

             x + 1 = 54 : 3

             x + 1 = 18

                   x = 18 -1 

                   x = 17

 b, 12x - 33 = 32.33

      12x - 33 =9 . 27

      12x - 33 = 243

             12x = 243 + 33

             12x = 276

               x=276:12

               x=23

a, 96 - 3 . ( x + 1 ) = 42

3 . ( x + 1 ) = 96 - 42

3 . ( x + 1 ) = 54

x + 1 = 54 : 3

x + 1 = 18

x = 18 - 1

x = 17

b, 12x - 33 = 3² . 3³

12x - 33 = 9 . 27

12x - 33 = 243

12x = 243 + 33

12x = 276

x = 276 : 12

x = 23

a)96-3.(x+1)=42

3.(x+1) = 96 - 42

3 . (x+1) = 54

x + 1 = 54 : 3

x + 1 = 18

x = 18-1

x = 17

a) \(\frac{x+3}{x+7}=\frac{96}{108}\)

\(\frac{x+3}{x+7}=\frac{8}{9}\)

\(8\left(x+7\right)=9\left(x+3\right)\)

\(8x+56=9x+27\)

\(9x-8x=56-27\)

\(x=29\)

b) \(\frac{4x+...}{-18}=-\frac{1}{6}\)(ĐỀ HÌNH NHƯ BỊ THIẾU)

\(6\left(4x+...\right)=\left(-1\right)\left(-18\right)\)

\(6\left(4x+...\right)=18\)

\(4x+...=18:6\)

\(4x+...=3\)

Bạn từ điền tiếp phần bị thiếu nhé

c) \(\frac{6}{3y-1}=\frac{33}{11}\)

\(\frac{6}{3y-1}=3\)

\(3y-1=\frac{6}{3}\)

\(3y-1=2\)

\(3y=2+1\)

\(3y=3\)

\(y=\frac{3}{3}=1\)

d) \(\frac{5z-3}{3}=\frac{28}{12}\)

\(\frac{5z-3}{3}=\frac{7}{3}\)

\(3\left(5z-3\right)=7.3\)

\(15z-9=21\)

\(15z=21+9\)

\(15z=30\)

\(z=\frac{30}{15}=2\)

Ths bạn nhìu nha

à , bạn ơi , chỗ bị thiếu là số 7 , mih viêt thiếu

(x+2)+(x+4)+...+(x+1996) = 99800

=> (x+x+x+...+x)+(2+4+...+1996) = 99800

=> 998x + 997002 = 99800

998x = 99800 - 997002

998x = -897202

x = -897202 : 998

x = -899

a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow x+100=0\)

hay x=-100

Mn nhớ giải chi tiết ra nha