a. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3$

$\Rightarrow x=2(-3)=-6; y=5(-3)=-15$

b. Áp dụng tính chất dãy tỉ số bằng nhau:

$7x=3y=\frac{x}{\frac{1}{7}}=\frac{y}{\frac{1}{3}}=\frac{x-y}{\frac{1}{7}-\frac{1}{3}}=\frac{16}{\frac{-4}{21}}=-84$

$\Rightarrow x=(-84):7=-12; y=-84:3=-28$

c. $\frac{x}{y}=\frac{5}{9}\Rightarrow \frac{x}{5}=\frac{y}{9}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{5}=\frac{y}{9}=\frac{3x}{15}=\frac{2y}{18}=\frac{3x+2y}{15+18}=\frac{66}{33}=2$

$\Rightarrow x=2.5=10; y=9.2=18$

d. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{15}=\frac{y}{7}=\frac{2y}{14}=\frac{x-2y}{15-14}=\frac{16}{1}=16$

$\Rightarrow x=16.15=240; y=7.16=112$

e.

Đặt $\frac{x}{5}=\frac{y}{2}=k\Rightarrow x=5k ; y=2k$

Khi đó: $xy=5k.2k=10k^2=1000\Rightarrow k^2=100\Rightarrow k=\pm 10$

Với $k=10$ thì $x=5k=50; y=2k=20$

Với $k=-10$ thì $x=5k=-50; y=2k=-20$

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{x}{13}-\frac{y}{7}-\frac{z}{5}=\frac{x-y-z}{13-7-5}=\frac{6}{1}=6$

$\Rightarrow x=13.6=78; y=7.6=42; z=5.6=30$

\(a.\Leftrightarrow\frac{5x^2+16}{\left(x+4\right)\left(x-4\right)}=\frac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}DKXD:x\ne4;-4\)

\(\Rightarrow5x^2+16=2x^2-8x-x+4+3x^2+12x-x-4\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

\(b.\Leftrightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}.DKXD:y\ne2;-2\)

\(\Rightarrow y^2+2y+y+2-5y+10=12+y^2-4\)

\(\Leftrightarrow-2y=-4\)

\(\Leftrightarrow y=2\)

a)

\(\dfrac{5x^2+16}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\) (\(x\ne\pm2\))

\(\Rightarrow\dfrac{5x^2+16}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Rightarrow\dfrac{5x^2+16-\left(2x^2-8x-x+4\right)-\left(3x^2+12x-x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)

\(\Rightarrow\dfrac{10x+16}{x^2-16}=0\)

=> 10x + 16 =0

=> 10x = -16

=> x = \(-\dfrac{8}{5}\)

\(G=\dfrac{1}{2}\left(\dfrac{x^{10}}{y^2}+\dfrac{y^{10}}{x^2}\right)+\dfrac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\)

\(=\dfrac{1}{2}\left(\dfrac{x^{10}}{y^2}+\dfrac{y^{10}}{x^2}\right)+\dfrac{1}{4}\left(x^{16}+y^{16}+1+1\right)-\left(1+x^2y^2\right)^2-\dfrac{1}{2}\)

\(\ge x^4y^4+x^4y^4-\dfrac{3}{2}-2x^2y^2-x^4y^4\)

\(=x^4y^4-2x^2y^2-\dfrac{3}{2}=\left(x^2y^2-1\right)^2-\dfrac{5}{2}\ge-\dfrac{5}{2}\)

Dấu = xảy ra khi: \(x^2=y^2=1\)

Theo Cô si:\(\dfrac{1}{2}\left(\dfrac{x^{10}}{y^2}+\dfrac{y^{10}}{x^2}\right)\ge\dfrac{1}{2}.2.\sqrt{x^8y^8}hay\ge x^4y^4\)

tương tự có \(\dfrac{1}{4}\left(x^{16}+y^{16}\right)\ge\dfrac{x^4y^4}{2}\)

Dấu = xảy ra ⇔ x= \(\pm y\)

Khi đó G = \(\dfrac{3}{2}x^4y^4-1-2x^2y^2-x^4y^4=\dfrac{1}{2}\left(x^4y^4-4x^2y^2+\text{4}\right)-3\)

G min = -3 khi \(x^4y^4-4x^2y^2+4=0\Leftrightarrow x^2y^2-2=0\) mà x=+-y suy ra x^4 =2 hay x=\(\pm\sqrt[4]{2}\)

Vậy có 4 cặp nghiệm thỏa mãn (x,y)=(\(\sqrt[4]{2},\sqrt[4]{2}\))\(\left(\sqrt[4]{2},-\sqrt[4]{2}\right),\left(-\sqrt[4]{2},\sqrt[4]{2}\right),\left(-\sqrt[4]{2},-\sqrt[4]{2}\right)\)

Đặt \(\left(x;y\right)=\left(\dfrac{1}{a};\dfrac{1}{b}\right)\)

BĐT trở thành: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}+\dfrac{16ab}{a+b}\ge5\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a^3+b^3}{ab}+\dfrac{16ab}{a+b}-5\left(a+b\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^3+b^3\right)+16a^2b^2-5ab\left(a+b\right)^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{ab\left(a+b\right)}\ge0\) (luôn đúng)

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

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow x^2=4.9=36\Rightarrow x=\sqrt{36}=6\\ y^2=4.16=64\Rightarrow y=\sqrt{64}=8\)

theo bài ra ta có:

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

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

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{x+16+y-25+z+9}{9+16+25}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{x+16}{9}=\dfrac{x+y+z}{50}\left(1\right)\)ta lại có:

\(\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\\ \Rightarrow\dfrac{7+2-x}{7}+\dfrac{9+2-x}{9}=2\\ \Rightarrow\left(1+\dfrac{2-x}{7}\right)+\left(1+\dfrac{2-x}{9}\right)=2\\ \Rightarrow\left(1+1\right)+\left(\dfrac{2-x}{7}+\dfrac{2-x}{9}\right)=2\\ \Rightarrow2+\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=2\\ \Rightarrow\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=0\\ \Rightarrow2-x=0\\ \Rightarrow x=2\)

thay x = 2 vào 1 ta có:

\(\Rightarrow\dfrac{2+16}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{18}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow2=\dfrac{x+y+z}{50}\\ \Rightarrow x+y+z=2.50\\ \Rightarrow x+y+z=100\)

vậy x + y + z = 100

Bạn thử nhân liên hợp cả tử và mẫu theo (A+B)×(A-B)=A^2 -B^2 xem s

\(a,\dfrac{y-1}{y+1}=\dfrac{\left(y-1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{y+1}{y-1}=\dfrac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{1}{y^2-1}=\dfrac{1}{\left(y-1\right)\left(y+1\right)}\\ b,\dfrac{2}{y^2-4y}=\dfrac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)};\dfrac{y}{y^2-16}=\dfrac{y^2}{y\left(y-4\right)\left(y+4\right)}\)