cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}vàa+b+c\ne0;a=2005.\)Tính b,c
Cho: \(\dfrac{a}{c}=\dfrac{a-b}{b-c},a\ne0,c\ne0,a-b\ne0,b-c\ne0\). CMR: \(\dfrac{1}{a}+\dfrac{1}{a-b}=\dfrac{1}{b-c}-\dfrac{1}{c}\)
\(A=\dfrac{a}{b+c}=\dfrac{c}{a+c}=\dfrac{b}{a+c}vàA=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}chứngminh(\dfrac{a+b+c}{b+c+d})^2=(\dfrac{a+b+c}{b+c+a})^2\)
Cho \(abc\ne0\) và \(a+b+c\ne0\). Tìm \(x\), biết: \(\dfrac{a+b-x}{c}+\dfrac{a+c-x}{b}+\dfrac{b+c-x}{a}+\dfrac{4x}{a+b+c}=1\)
Lời giải:
PT $\Leftrightarrow \frac{a+b-x}{c}+1+\frac{a+c-x}{b}+1+\frac{b+c-x}{a}+1+\frac{4x}{a+b+c}-4=0$
$\Leftrightarrow \frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4(a+b+c-x)}{a+b+c}=0$
$\Leftrightarrow (a+b+c-x)(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c})=0$
$\Rightarrow a+b+c-x=0$ hoặc $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$, khi đó $x$ nhận mọi giá trị thực.
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\neq 0$
$\Rightarrow a+b+c-x=0$
$\Rightarrow x=a+b+c$
Cho \(b\ne-d;b\ne-3d;b\ne0;d\ne0\) và \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\) . Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)
Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)
+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{2c}{2d}=\dfrac{c}{d}\) (1)
\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{3a+3c}{3b+3d}=\dfrac{a+3c-\left(3a+3c\right)}{b+3d-\left(3b+3d\right)}=\dfrac{-2a}{-2b}=\dfrac{a}{b}\) (2)
(1);(2) \(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Cho \(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{1}{a+b+c}\left(a,b,c\ne0,a+b+c\ne0\right)\)
Tính \(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)\)
Ai giúp mik đi, mik cho 5 coin
\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)
Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)
Từ đó tự giải ra
Áp dụng t/c dtsbn:
\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{b+c-5+a+c+2+a+b+3}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\)\(\left(1\right)\)
Mặt khác \(\dfrac{1}{a+b+c}=\dfrac{b+c-5}{a}=2\)\(\Rightarrow a+b+c=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{2}-c\\a+c=\dfrac{1}{2}-b\\b+c=\dfrac{1}{2}-a\end{matrix}\right.\)\(\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-a-5=2a\\\dfrac{1}{2}-b+2=2b\\\dfrac{1}{2}-c+3=2c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\b=\dfrac{5}{6}\\c=\dfrac{7}{6}\end{matrix}\right.\)
\(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)=\left(-\dfrac{3}{2}-3.\dfrac{5}{6}\right)\left(\dfrac{5}{6}-\dfrac{7}{6}\right)\left(3.\dfrac{7}{6}+\dfrac{3}{2}\right)=\dfrac{20}{3}\)
Cho \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(a;b;c\ne0;b\ne c\right).\) Chứng minh: \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Cho \(abc\ne0\) và \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}.\) Tính \(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Giúp ik
Lời giải:
Bạn tham khảo cách làm tương tự tại đây:
https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584
Kết quả $P=8$ hoặc $P=-1$
Cho \(a,b,c\ne0\) và \(a+b+c=\dfrac{a+2b-c}{c}=\dfrac{b+2c-a}{a}=\dfrac{c+2a-b}{b}\)
Tính \(P=\left(2+\dfrac{a}{b}\right)\left(2+\dfrac{b}{c}\right)\left(2+\dfrac{c}{a}\right)\)
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e, \(\dfrac{x+5}{2}=\dfrac{y-2}{3}vàx-y=10\)
f, \(\dfrac{a+2}{3}=\dfrac{b-7}{5}vàa-b+c=-33\)
h,\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}và5a-3b-4c=500\)
Zúp mìk zới!
e: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)
=>x+5=-34; y-2=-51
=>x=-39; y=-49
g: Áp dụng tính chất của DTSBN, ta được
\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)
=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13
=>a=-493/13; b=-1051/13; c=-1453/13
Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$
Suy ra:
$x+5=2(-17)=-34\Rightarrow x=-39$
$y-2=3(-17)=-51\Rightarrow y=-49$
f. Đề thiếu. Bạn xem lại
h. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$
$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$
$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$
$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$
Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$
$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$
$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$