b)

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{3x}{15}=\dfrac{2y}{6}=\dfrac{7z}{14}=\dfrac{3x-2y+7z}{15-6+14}=\dfrac{69}{23}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=15\\y=9\\z=6\end{matrix}\right.\)

c)

\(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{9}\\y=\dfrac{4}{9}\end{matrix}\right.\)

d)

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{18}{-2}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}x=-9.7=-63\\y=-9.5=-45\end{matrix}\right.\)

a)

\(\dfrac{x}{5}=\dfrac{y}{7}=k\\ \Rightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)

\(x.y=5k.7k=35k^2=140\\ \Rightarrow k^2=4\Rightarrow k=\pm2\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=10\\y=14\end{matrix}\right.\\\left\{{}\begin{matrix}x=-10\\y=-14\end{matrix}\right.\end{matrix}\right.\)

bữa sau 1 là bạn đánh câu hỏi, 2 là bạn đăng hình lớn lên

hình kiểu này thì đọc cho lời mắt -_-

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)

\(B=x^2-6x+2004\\ B=x^2-6x+9+1995\\ B=\left(x-3\right)^2+1995\ge1995\)

đẳng thức xảy ra khi x-3=0 => x=3

vậy MIN_B=1995 tại x=3

\(C=4x^2+4x+2018\\ C=4x^2+4x+1+2017\\ C=\left(2x+1\right)^2+2017\ge2017\)

đẳng thức xảy ra khi 2x+1=0 => x=-1/2

vậy MIN_C=2017 tại x=-1/2

\(\sqrt{x^2+12}+5=3x+\sqrt{x^2+5}\\ \Leftrightarrow\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\\ \Leftrightarrow\left(\sqrt{x^2+12}-\sqrt{x^2+5}\right)=\left(3x-5\right)^2\\ \Leftrightarrow2x^2+17-2\sqrt{\left(x^2+12\right)\left(x^2+5\right)}=9x^2-30x+25\\ 7x^2-30x+8=-2\sqrt{\left(x^2+12\right)\left(x^2+5\right)}\\ \Leftrightarrow\left(7x^2-30x+8\right)^2=\left(-2\sqrt{\left(x^2+12\right)\left(x^2+5\right)}\right)^2\\ \Leftrightarrow49x^4-420x^3+1012x^2-480x+64=4x^4+68x^2+240\\ \Leftrightarrow45x^4-420x^3+944x^2-480x-176=0\Leftrightarrow\left(x-2\right)\left(45x^3-330x^2+284x+88\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\45x^3-330x^2+284x+88=0\end{matrix}\right.\Rightarrow x=2\)

cái phương trình thứ 2 mình ko bt giải, thông cảm nha :))

\(A=1+2+2^2+2^3+...+2^{99}+2^{100}\\ 2A=2+2^2+2^3+...+2^{100}+2^{101}\\ 2A-A=\left(2+2^2+2^3+...+2^{100}+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{99}+2^{100}\right)\\ A=2^{101}-1\)

a)fix đề : \(x^2\left(x-2\right)+x-2=0\)

\(x^2\left(x-2\right)+x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

b)

\(5\left(x-11\right)-x+11=0\\ \Leftrightarrow\left(x-11\right)\left(5-1\right)=0\\ \Leftrightarrow x-11=0\Leftrightarrow x=11\)