bạn giải đi bạn

Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:

\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)

\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)

1/

\(\frac{x}{y}=\frac{3}{4}\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{4x}{12}=\frac{5y}{20}=\frac{4x-5y}{-8}\) (1)

\(\frac{x}{3}=\frac{y}{4}=\frac{3x}{9}=\frac{4y}{16}=\frac{3x+4y}{25}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{4x-5y}{-8}=\frac{3x+4y}{25}\Rightarrow\frac{4x-5y}{3x+4y}=\frac{-8}{25}\)

2/

\(M-N=3x\left(x-y\right)-\left(y-x\right)\left(y+x\right)=\)

\(=3x\left(x-y\right)+\left(x-y\right)\left(y+x\right)=\left(x-y\right)\left(4x+y\right)\)

Mà \(x-y\) chia hết cho 11 nên \(M-N\) chia hết cho 11

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

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)

=> \(\hept{\begin{cases}\frac{x}{2}=9\\\frac{y}{4}=9\\\frac{z}{-4}=9\end{cases}}\)  =>   \(\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)

Vậy ...

a, ÁP DỤNG DÃY TỈ SỐ BĂNG NHAU TA CÓ

\(\frac{x}{2}=\frac{y}{3}=\frac{x}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)

\(\Rightarrow\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)

\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)

\(C=\frac{2x+3y}{3x+4y}=\frac{2\cdot\frac{3}{4y}+3y}{3\cdot\frac{3y}{4}\cdot4y}\)

\(=\frac{2\cdot\frac{3}{4}+3}{3\cdot\frac{3}{4}+4}=\frac{\frac{9}{2}}{\frac{25}{4}}\)

\(=\frac{9}{2}\cdot\frac{4}{25}=\frac{18}{25}\)

\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)

\(\Rightarrow C=\frac{\frac{3y}{2}+3y}{\frac{9y}{4}+4y}=\frac{\frac{9y}{2}}{\frac{25y}{4}}=\frac{18}{25}\)

a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)

Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)

Q = \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Q = \(\frac{x^2-2x-x+2+4x+4-3x-6}{\left(x+2\right)\left(x-2\right)}\)

Q = \(\frac{x^2-2x}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)

b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3

Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)

=> x + 2 = 3(x - 3)

=> x + 2 = 3x - 9

=> x - 3x = -9 - 2

=> -2x = -11

=> x = 11/2 (tm)

Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)

c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3

Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)

Để M \(\in\)Z <=> 3 \(⋮\)x - 3

=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng:

Vậy ...

a) ta có: \(\frac{x}{y}=\frac{3}{4}\Rightarrow4x=3y\)

\(D=\frac{4x-5y}{3x+4y}=\frac{3y-5y}{3y+4y-x}=\frac{-2y}{7y-x}=\frac{-2y}{7y-y3:4}\)

\(=\frac{-2y}{\frac{25}{4}y}=-2y:\left(\frac{25}{4}y\right)=-\frac{8}{25}\)

b) ta có: M=3x.(x-y) chia hết cho 11

N = y² - x² = y² - xy - x² + xy = y.(y-x) - x.(x-y) = (y-x).(y+x) = - (x-y).(y+x) chia hết cho 11

=> M-N chia hết cho 11 (đpcm)

mn ơi nhanh đi a

\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)

\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)

\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

\(\Rightarrow x=\frac{16}{11}\)