\(a,\left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x=5\\ \Rightarrow-2x=-4\\ \Rightarrow x=2\\ b,x\left(x-6\right)+2x-12=0\\ \Rightarrow x\left(x-6\right)+2\left(x-6\right)=0\\ \Rightarrow\left(x+2\right)\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)

Tính (theo mẫu).

`a) 2 cm xx 5 =10cm `               

`10 cm : 5 = 2cm`

`b) 2 kg xx 3=6kg   `               

`6 kg : 3=2kg`

`c) 2 l xx 4=8l`                     

` 8l : 4 = 2l`

a) 2 cm x 5 = 10 cm

10 cm : 5 = 2 cm

b) 2 kg x 3 = 6 kg

6 kg : 3 = 2 kg

c) 2 l x 4 = 8 l

8 l : 4 = 2 l

b,2kg x 3 =6kg

6kg : 3 = 2kg

c,2l x 4 = 8l

8l :4 = 2l

a) \(x+\dfrac{1}{2}=\dfrac{3}{5}\)

    \(x=\dfrac{3}{5}-\dfrac{1}{2}\) 

    \(x=\dfrac{6}{10}-\dfrac{5}{10}\)

    \(x=\dfrac{1}{10}\)

b) \(x\times\dfrac{6}{7}=\dfrac{5}{21}\)

    \(x=\dfrac{5}{21}:\dfrac{6}{7}\)

    \(x=\dfrac{5}{21}\times\dfrac{7}{6}\)

    \(x=\dfrac{5}{18}\)

c) \(x:\dfrac{4}{3}=\dfrac{8}{9}\)

    \(x=\dfrac{8}{9}\times\dfrac{4}{3}\)

   \(x=\dfrac{32}{27}\)

a: x=3/5-1/2=6/10-5/10=1/10

b: x=5/21:6/7=5/21x7/5=35/105=1/3

c: x=8/9x4/3=32/27

\(a.x+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{10}\)

\(b.x\times\dfrac{6}{7}=\dfrac{5}{21}\)

\(x=\dfrac{5}{21}:\dfrac{6}{7}=\dfrac{5}{18}\)

\(c.x:\dfrac{4}{3}=\dfrac{8}{9}\)

\(x=\dfrac{8}{9}\times\dfrac{4}{3}=\dfrac{32}{27}\)

\(a,\)Với \(x\ne-3,x\ne2\) ta có :

\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)

   \(=\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

   \(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

  \(=\dfrac{x-4}{x-2}\)

\(b,\) \(A=-3\Leftrightarrow\dfrac{x-4}{x-2}=-3\)

\(\Leftrightarrow x-4=-3\left(x-2\right)\)

\(\Leftrightarrow x-4+3x-6=0\)

\(\Leftrightarrow4x=10\Rightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)

c) Để A đạt giá trị nguyên dương thì \(\left\{{}\begin{matrix}x-4⋮x-2\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2⋮x-2\\x>2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\inƯ\left(-2\right)\\x>2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2\in\left\{1;-1;2;-2\right\}\\x>2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;1;4;0\right\}\\x>2\end{matrix}\right.\Leftrightarrow x=4\)

Vậy: Để A là số nguyên dương thì x=4

a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)

a ,   x = 5 + a b ,   x = − 10 − b

a: Thay x=-1 và y=5 vào y=ax+6, ta được:

6-x=5

hay x=1

b: Vì đồ thị hàm số y=ax+b đi qua hai điểm (1;1) và (0;-2) nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}a+b=1\\b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1-b=1-\left(-2\right)=1+2=3\\b=-2\end{matrix}\right.\)

a) x = 10 ; y = 8

Ta có : 4x = 5y => \(\frac{x}{5}=\frac{y}{4}\)

Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)

=> xy = 5k.4k = 20k²

=> 20k² = 80

=> k² = 4 => k = \(\pm2\)

Với k = 2 thì x = 5.2 = 10 , y = 4.2 = 8

Với k = -2 thì x = 5.(-2) = -10 , y = 4(-2) = -8

b) Ta có : \(2a=5b=3c\)=> \(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{5}}=\frac{c}{\frac{1}{3}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{5}}=\frac{c}{\frac{1}{3}}=\frac{a+b-c}{\frac{1}{2}+\frac{1}{5}-\frac{1}{3}}=\frac{-44}{\frac{11}{30}}=-120\)

Từ đó suy ra a = -60,b = -24,c = -40

a. Ta có : \(4x=5y\Rightarrow x=\frac{5}{4}y\)

Mà xy = 80

\(\Rightarrow\frac{5}{4}y.y=80\)

\(\Rightarrow y^2=64\)

\(\Rightarrow y^2=8^2\)

=> y = 8 hoặc y = - 8

+) y = 8 => x = 80 : y = 80 : 8 = 10

+) y = - 8 => x = 80 : ( - 8 ) = - 10

Vậy các cặp ( x ; y ) thỏa mãn đề bài là : ( 10 ; 8 ) ; ( - 10 ; - 8 )  

b. \(2a=5b=3c\Rightarrow\frac{2a}{30}=\frac{5b}{30}=\frac{3c}{30}\Rightarrow\frac{a}{15}=\frac{b}{6}=\frac{c}{10}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có :

\(\frac{a}{15}=\frac{b}{6}=\frac{c}{10}=\frac{a+b-c}{15+6-10}=\frac{-44}{11}=-4\)

Suy ra :

+) \(\frac{a}{15}=-4\Leftrightarrow a=-60\)

+) \(\frac{b}{6}=-4\Leftrightarrow b=-24\)

+) \(\frac{c}{10}=-4\Leftrightarrow c=-40\)

a) Tìm x

\(6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{\left(-2\right)^{2012}}\Rightarrow6-\left(x-\frac{1}{3}\right)^2=\frac{2^{2013}}{2^{2012}}=2^1=2\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^2=6-2=4=2^2\Rightarrow\hept{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}}\)

Vậy \(x\in\left\{\frac{7}{3};\frac{-5}{3}\right\}\)

b) Ta có : \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\) và \(5b=7c\Rightarrow\frac{b}{7}=\frac{c}{5}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{a}{21}=\frac{b}{14}\\\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{b}{14}=\frac{c}{10}\end{cases}}\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\) 

Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{a+b-c}{21+14-10}=-\frac{50}{25}=-2\)

\(\Rightarrow a=\left(-2\right).21=-42\)     \(b=\left(-2\right).14=-28\)     \(c=\left(-2\right).5=-10\)

Vậy a = -42 ; b = -28 và c = -10

a ,   x ∈ 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; ...... b ,   x ∈ 0 ; 1 ; 2 ; 3 ; 4 ; 5