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Ta có: \(a+b+c=1 \)
\(\Leftrightarrow(a+b+c)^2=1 \)
\(\Leftrightarrow ab+bc+ca=0 (1) \)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{(x+y+z)}{\left(a+b+c\right)}=x+y+z\)
\(\Leftrightarrow x=a\left(x+y+z\right)\)
\(\Leftrightarrow y=b.\left(x+y+z\right)\)
\(\Leftrightarrow z=c.\left(x+y+z\right)\)
\(\Rightarrow xy+yz+zx=ab.\left(x+y+z\right)^2+bc.\left(x+y+z\right)^2+ca.\left(x+y+z\right)^2\)
\(\Leftrightarrow xy+yz+zx=\left(ab+bc+ca\right).\left(x+y+z\right)^2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra: \(xy+yz+zx=0\)
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)
\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)
\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)
\(\Rightarrow\left(a+b+c\right)^2=60\)
\(\Rightarrow a+b+c=\pm\sqrt{60}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)
Các câu sau làm tương tự
b. \(ab=\dfrac{3}{5};bc=\dfrac{4}{5};ac=\dfrac{3}{4}\)
\(\Rightarrow ab\cdot bc\cdot ac=\dfrac{9}{25}\Rightarrow\left(abc\right)^2=\dfrac{9}{25}\Rightarrow abc=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\dfrac{3}{5}:bc=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\\b=\dfrac{3}{5}:ac=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\c=\dfrac{3}{5}:ab=\dfrac{3}{5}:\dfrac{3}{5}=1\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\dfrac{3}{5}:\dfrac{4}{5}=-\dfrac{3}{4}\\b=-\dfrac{3}{5}:\dfrac{3}{4}=-\dfrac{4}{5}\\c=-\dfrac{3}{5}:\dfrac{3}{5}=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy......................
a; \(\dfrac{1}{2}-\dfrac{-3}{6}+\dfrac{5}{3}-\dfrac{9}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{5}{3}-\dfrac{3}{4}\)
\(=1-\dfrac{3}{4}+\dfrac{5}{3}=\dfrac{1}{4}+\dfrac{5}{3}=\dfrac{3+20}{12}=\dfrac{23}{12}\)
b: \(=\dfrac{3}{11}\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}\cdot\dfrac{-6-16}{9}=\dfrac{3}{11}\cdot\dfrac{-22}{9}=\dfrac{-2}{3}\)
c: \(=1-3+\dfrac{1}{4}=-2+\dfrac{1}{4}=-\dfrac{7}{4}\)
Đề ảo tek.Sửa đề.
\(\left\{{}\begin{matrix}a+b+c=5\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=25\\\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\bc+ac+ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\2bc+2ac+2ab=0\end{matrix}\right.\)
\(\Leftrightarrow a^2+b^2+c^2+2ab-2ab+2bc-2bc+2ac-2ac=25\)
\(\Leftrightarrow a^2+b^2+c^2=25\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
c: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)
Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b.\(ĐK:x;y\in Z^+;x;y\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)
\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)
\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)
\(\Leftrightarrow x=\dfrac{5y}{y-5}\)
\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )
Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
TH1:
\(y-5=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm ) ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )
Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:
\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)
Câu a mik ko bt nên bạn tham khảo nhé:
https://hoc24.vn/cau-hoi/cho-a-b-c-0-va-day-ti-so-dfrac2bc-aadfrac2c-babdfrac2ab-cctinh-p-dfracleft3a-2brightleft3b-2crightleft.177725456910
a) Ta co: a/b = c/d= k
=> a=bk
c=dk
Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)
Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)
Tu (1) va (2)
=> a-b/a+b=c-d/c+d
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b) Từ (*) ta có:
\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)
Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)
\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)
Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2b+c-a}{a}=\frac{2c+a-b}{b}=\frac{2a+b-c}{c}=\frac{2b+c-a+2c+a-b+2a+b-c}{a+b+c}\)
\(=\frac{2(a+b+c)}{a+b+c}=2\)
Do đó: \(\left\{\begin{matrix} 2b+c-a=2a\\ 2c+a-b=2b\\ 2a+b-c=2c\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 2b=3a-c\\ 2c=3b-a\\ 2a=3c-b\end{matrix}\right.\) và \(\left\{\begin{matrix} c=3a-2b\\ a=3b-2c\\ b=3c-2a\end{matrix}\right.\)
Suy ra: \(P=\frac{(3a-2b)(3b-2c)(3c-2a)}{(3a-c)(3b-a)(3c-b)}=\frac{c.a.b}{2b.2c.2a}=\frac{1}{8}\)
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)<=>\(\dfrac{2b+c}{a}-1=\dfrac{2c+a}{b}-1=\dfrac{2a+b}{c}-1\)
<=>\(\dfrac{2b+c}{a}=\dfrac{2c+a}{b}=\dfrac{2a+b}{c}=\dfrac{2b+c+2c+a+2a+b}{a+b+c}=\dfrac{3\left(a+b+c\right)}{a+b+c}=3\)=>\(\left\{{}\begin{matrix}2b+c=3a\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3a-c=2b\end{matrix}\right.\\2c+a=3b\Rightarrow\left\{{}\begin{matrix}3b-2c=a\\3b-a=2c\end{matrix}\right.\\2a+b=3c\Rightarrow\left\{{}\begin{matrix}3c-2a=b\\3c-b=2a\end{matrix}\right.\end{matrix}\right.\) thay vào
\(P=\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}=\dfrac{c.a.b}{2b.2c.2a}=\dfrac{1}{8}\)
Giải:
Có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b.\left(3k+1\right)}=\dfrac{c}{d.\left(3k+1\right)}\)
\(\Leftrightarrow\dfrac{a}{3bk+b}=\dfrac{c}{3dk+d}\)
\(\Leftrightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\) (Vì \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\))
\(\Leftrightarrowđpcm\)
Chúc bạn học tốt!
Có \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Thay (1) vào \(\dfrac{a}{3a+b}\)
\(\Rightarrow\)\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}\)
\(=\dfrac{k}{3k+1}\) (2)
Thay (1) vào \(\dfrac{c}{3c+d}\)
\(\Rightarrow\)\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}\)
\(=\dfrac{k}{3k+1}\) (3)
Từ (2) và (3)
=> đpcm