( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0

 <=> +) 1/7 . x - 2/7 = 0                                    +)    (- 1 / 5) . x +3/5 = 0                              +)  1/ 3 . x + 4/ 3 = 0

                    x = 2                                                                  x = 3                                                         x = 4

                                                    Vậy x = 2 : x = 3 ; x=4

\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)

\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)

\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)

\(A=2x^4y^4z^2\)

\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)

\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)

\(B=8x^7y^{y^8}z^6\)

\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)

\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)

\(\Leftrightarrow-16x=-14\)

\(\Rightarrow x=\dfrac{7}{8}\)

\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)

\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)

\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)

Cái này không biết tách kiểu gì cho vừa nên bạn nhấn máy tính nhé

Mode 5 3 rồi lần lượt điền vào theo thứ tự trên thì

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)

a) Đặt \(x^2+3x+1=y\) khi đó ta có:

\(y\left(y-4\right)-5\)

\(=y^2-4y-5\)

\(=y\left(y-5\right)+\left(y-5\right)\)

\(=\left(y+1\right)\left(y-5\right)\)

Thay \(y=x^2+3x+1\):

\(\left(x^2+3x+1+1\right)\left(x^2+3x+1-5\right)\)

\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)

\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)

\(=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x+4\right)\)

b) Biến đổi 3 số sau có chứa x² + 2x rồi đặt ẩn.

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

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

Đặt \(x^2+8x+7=y'\)

Khi đó ta đc:

\(y'\left(y'+8\right)+15\)

\(=\left(y'\right)^2+8y'+15\)

\(=y'\left(y'+3\right)+5\left(y'+3\right)\)

\(=\left(y'+5\right)\left(y'+3\right)\)

....

d) \(x^2-2xy+y^2-7x+7y+12\)

Biến đổi chứa x - y rồi đặt ẩn.

a ) \(x+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)

\(x\times1+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)

\(x\times1+x\times\frac{1}{3}\times\frac{9}{2}+x\times\frac{7}{2}=252\)

\(x\times1+x\times\frac{9}{6}+x\times\frac{7}{2}=252\)

\(x\times\left(1+\frac{9}{6}+\frac{7}{2}\right)=252\)

\(x\times6=252\)

\(x=252\div6\)

\(x=42\)

b ) \(\left(3\times x-0,8\right)\div x+14,5=15\)

\(\left(3\times x-0,8\right)\div x=15-14,5\)

\(\left(3\times x-0,8\right)\div x=0,5\)

\(0,8\div x=3-0,5\)

\(0,8\div x=2,5\)

\(x=0,8\div2,5\)

\(x=0,32\)

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a= 1/10

b= 18/5

c= -49/8

d= 413

.