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Trần Thị Hảo
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.
\(\left\{{}\begin{matrix}a+b+c=5\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)^2=25\\\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\bc+ac+ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2+c^2+2ab+2bc+2ac=25\\2bc+2ac+2ab=0\end{matrix}\right.\)
\(\Leftrightarrow a^2+b^2+c^2+2ab-2ab+2bc-2bc+2ac-2ac=25\)
\(\Leftrightarrow a^2+b^2+c^2=25\)
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
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: \(xy+yz+zx=0\)
a: f(x)=|5x-4|
b: f(x)=6
=>|5x-4|=6
=>5x-4=6 hoặc 5x-4=-6
=>5x=10 hoặc 5x=-2
=>x=2 hoặc x=-2/5
Áp dụng định lí tổng ba góc trong một tam giác, có:
A+B+C=180 độ
=> A+B=180-C
Mặt khác: A-B=50 độ (gt) (*)
=> A= \(\dfrac{\left[\left(A+B\right)+\left(A-B\right)\right]}{2}\)
\(\Rightarrow\)\(A=\left(180-C+50\right).\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{2}C=\left(230-C\right).\dfrac{1}{2}\)
\(\Rightarrow C=230-C\)
\(\Rightarrow2C=230\)
\(\Rightarrow C=115\) độ
Do đó: A=\(\dfrac{1}{2}.115\)
=>A=\(\dfrac{115}{2}\)
Thay A=\(\dfrac{115}{2}\)vào (*), có:
\(\dfrac{115}{2}-B=50\)
\(\Rightarrow B=\dfrac{115}{2}-50\)
\(\Rightarrow B=\dfrac{15}{2}\)
Vậy....
(Mik ko bt số đo của một góc có thể là phân số hay ko. Nhưng cách làm thì chắc đúng)
\(\left\{{}\begin{matrix}\widehat{A}-\widehat{B}=50^o\\\widehat{A}=\dfrac{1}{2}\widehat{C}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{A}-50^o\\\widehat{C}=2\widehat{A}\end{matrix}\right.\)
Thay vào ta có:
\(\widehat{A}+\left(\widehat{A}-50^o\right)+2\widehat{A}=180^o\)
\(\Rightarrow4\widehat{A}-50^o=180^o\)
\(\Rightarrow4\widehat{A}=230^o\Leftrightarrow\widehat{A}=\dfrac{230^o}{4}=57,5^o\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=57,5^o-50^o=7,5^o\\\widehat{C}=57,5^o.2=115^o\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)
\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)
\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)
\(\Rightarrow\left(a+b+c\right)^2=60\)
\(\Rightarrow a+b+c=\pm\sqrt{60}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)
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......................