Viết đề bài rõ ra

(mười trừ hai nhân X)nhân(mười hai trừ ba nhân X)

Đề bài:

(10-2x)(12-3x)=0

Trả lời:

(10-2x)(12-3x)=0

=>10-2x=0

hoặc 12-3x=0

=>x=5

hoặc x=4

3==D(Koo)3==D

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

\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)

\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)

b/ x + (x + 1) + (x + 2) + .... + (x + 30) = 1240 

 x + (x + x + ....+ x) + (1 + 2 + ... + 30) = 1240 

x + 30x + 465 = 1240 

31x = 1240 - 465 

31x = 775 

=> x = 775 : 31 = 25 

c/ Sửa đề (x + 4) + (x + 6) + (x + 8) + .... + (x + 26) = 210 

 (x + x + x + .... + x) + (4 + 6 + 8 + .... + 26) = 210 

 12x + 180 = 210 

12x = 210 - 180 = 30 

=> x = 30 : 12 = 2,5 

Cảm ơn bạn nha

a) x + 2x + 3x + 4x + ... + 19x = 570

=> ( 1 + 2 + 3 + ... + 19 )x = 570

=> [( 19 + 1 ) . 19 : 2 ]x = 570

=> 190x = 570

=> x = 3

lớp 1 cái con khỉ gió

\(2\times x+3\times x=15\)

\(\left(2+3\right)\times x=15\)

\(5\times x=15\)

\(x=15\div5\)

\(x=3\)

x=3

h mik nha

\(a,=2x\left(x+3\right)\\ b,=x^3\left(x+3\right)+\left(x+3\right)=\left(x^3+1\right)\left(x+3\right)\\ =\left(x+1\right)\left(x+3\right)\left(x^2-x+1\right)\\ c,=64-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\\ A=x^2+6x+5+x^3-8-x^2-x+2\\ A=x^3+5x-1\)

a) 2x²+6x=2x(x+3)
b) x⁴+3x³+x+3=(x⁴+x)+(3x³+3)=x(x³+1)+3(x³+1)=(x+3)(x³+1)
c) 64-x²-y²+2xy=-(x²-2xy+y²)+8²=8-(x+y)²=(8+x+y)(8-x-y)

A= (x+5)(x+1)+(x-2)(x²+2xx+4)-(x²+x-2)
A= x²+6x+5+x³-8-x²-x+2
A= x³+(x²-x²)+(6x-x)+(5-8+2)
A= x³+5x-1

a)\(\frac{2}{3}\cdot x+\frac{1}{4}=\frac{5}{6}\)

\(\frac{2}{3}\cdot x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac{2}{3}=\frac{7}{8}\)

b) \(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{1}{4}:\frac{2}{3}\)

\(\frac{7}{12}-\frac{5}{6}\cdot x=\frac{3}{8}\)

\(\frac{5}{6}\cdot x=\frac{7}{12}-\frac{3}{8}=\frac{5}{24}\)

\(x=\frac{5}{24}:\frac{5}{6}=\frac{1}{4}\)

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

\(\frac{7}{3}-\left(x+1\right)=\frac{5}{9}\)

\(x+1=\frac{7}{3}-\frac{5}{9}=\frac{16}{9}\)

\(x=\frac{16}{9}-1=\frac{7}{9}\)

d) \(\frac{2\cdot x+1}{15}=\frac{3}{5}\)

\(\left(2\cdot x+1\right):15=\frac{3}{5}\)

\(2\cdot x+1=\frac{3}{5}\cdot15=9\)

\(2\cdot x=9-1=8\)

\(x=8:2=4\)

a, 2/3 . x +1/4=5/6

      2/3 . x=5/6-1/4

      2/3 . x=10/12 -3/12

       2/3 . x=7/12

       x= 7/12 : 2/3

      x=7/8 

Vậy x=7/8

số hạng thứ 100 là 100x+199

x=10

(x+1)(6x²+2x)+(x-1)(6x²+2x)
<=> (6x²+2x)(x+1+x-1)
<=> 2x(3x+1)2x
<=> 4x²(3x+1)
<=> x²=0
       3x+1=0
<=> x=0
       x= -1/3 (-1 phần 3)

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

Thay x,y vào ta có :

A=[2.2+3.(-5)]-[3.2-5.(-5)]

A=[4+(-15)]-[6-(-25)]

A=(-11)-31

A=20

Thanks nhìu!