Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z+3}{-4}=\frac{2x-2+3y-6-z+3}{9}=\frac{56-5}{9}\)\(=\frac{17}{3}\)
\(\Rightarrow x=\frac{37}{3},y=19,z=\frac{77}{3}\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\); \(2x+3y-z=56\)
\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4};2x+3y-z=56\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{56-2-6+3}{9}=\frac{51}{9}=\frac{17}{3}\)
\(\Leftrightarrow x=\frac{37}{3};y=19;z=\frac{77}{3}\)
Vậy \(x=\frac{37}{3};y=19;z=\frac{77}{3}\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)
\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)
\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right).x^2.\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
Ta có:
\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)x^2\left(1-2x\right)\)
\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)
\(=\left(x-2\right)\left[\left(2x^3-x^2+1\right)+\left(x^2-2x^3\right)\right]\)
\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)
\(=\left(x-2\right).1\)
\(=x-2\)
\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
1. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+2}{3}=\frac{y-7}{5}=\frac{x+y-5}{3+5}=\frac{16}{8}=2\Rightarrow\hept{\begin{cases}x+2=6\\y-7=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=17\end{cases}}}\)
2. áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-y+2}{2-3}=\frac{-10+7}{-1}=3\Rightarrow\hept{\begin{cases}x+5=6\\y-2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=11\end{cases}}\)
\(2^x+3\cdot2^{x+1}=56\)
=>\(2^x+6\cdot2^x=56\)
=>\(7\cdot2^x=56\)
=>\(2^x=8\)
=>x=3