a: -2x(x+3)+x(2x-1)=10

=>-2x^2-6x+2x^2-x=10

=>-7x=10

=>x=-10/7

b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3

=>3x^2+1/6x-3x^2-2=3

=>1/6x-2=3

=>x=30

\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)

\(\Leftrightarrow-24x+27=10\)

\(\Leftrightarrow-24x=-27\)

\(\Leftrightarrow x=\dfrac{27}{24}\)

Vậy \(x=\dfrac{27}{24}\)

\(b,x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)

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

Vậy \(x\in\left\{0;\pm5\right\}\)

2(x-1)-3(2x+2)-4(2x+3)=16

2x-2-6x+6-8x+12=16

(2x-6x-8x)-(2+6+12)=16

(-12).x-20=16

(-12).x=16+20=36

x=36:(-12)=-3

2(x-1)-3(2x+2)-4(2x+3)=16

2x-2-6x-6-8x-12=16

2x-6x-8x=16+2+6+12

-12x=36

x=-3

Bạn ấy trả lời đúng rồi

Mike bạn ấy nhé

Chúc học tốt

a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

áp dụng tính chất dãy tỉ số = nhau

\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)

\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)

\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)

\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)

b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)

có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)

\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)

áp dụng t/c dãy tỉ số = nhau

\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)

\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)

\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)

\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)

c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)

thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(=>y=2\)

\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)

d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)

thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)

\(=>x=\dfrac{2.3}{3}=2\)

(x+3).(y+1)=3

--->x+3,y+1 thuộc Ư(3)={1,3,-1,-3}

Ta có bảng sau

x+3               1                           -1 

y+1               3                           -3

y                   2                           -4

x                   -2                          -4

--->(x,y) thuộc(-2,2),(-4,-4)