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1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
1.
a.
\(\left(\dfrac{-4}{5}+\dfrac{2}{3}\right)\cdot\dfrac{7}{11}+\left(\dfrac{-1}{5}+\dfrac{1}{3}\right)\cdot\dfrac{7}{11}\\ =\dfrac{7}{11}\cdot\left(\dfrac{-4}{5}+\dfrac{2}{3}+\dfrac{-1}{5}+\dfrac{1}{3}\right) \\ =\dfrac{7}{11}\cdot\left[\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\\ =\dfrac{7}{11}\cdot\left[\left(-1\right)+1\right]\\ =\dfrac{7}{11}\cdot0\\ =0\)
b.
\(\left(-3^2\right)\cdot\left(\dfrac{3}{4}-0,25\right)-\left|-2\right|\\ =\left(-9\right)\cdot0,5-2\\ =-4,5-2\\ =-6,5\)
2.
\(y=f\left(x\right)=\left(m+1\right)x\\ \Rightarrow4=f\left(2\right)=\left(m+1\right)\cdot2\\ \Rightarrow m+1=2\\ \Leftrightarrow m=1\)
Tự
3.
a.
\(\left|x-\dfrac{2}{5}\right|=\dfrac{3}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{3}{4}\\x-\dfrac{2}{5}=\dfrac{-3}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{-7}{20}\end{matrix}\right.\)
b.
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}=\dfrac{x+2y-z}{5+6-4}=\dfrac{14}{7}=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\\y=6\\z=8\end{matrix}\right.\)
Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
1,
a, \(\left(x-\dfrac{1}{7}\right)^4=\left(x-\dfrac{1}{7}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{7}\right)^4-\left(x-\dfrac{1}{7}\right)^2=0\)
\(\Leftrightarrow\left[\left(x-\dfrac{1}{7}\right)^2+x-\dfrac{1}{7}\right]\left[\left(x-\dfrac{1}{7}\right)^2-x+\dfrac{1}{7}\right]=0\)
\(\Leftrightarrow\left[x^2+\dfrac{1}{49}-\dfrac{2}{7}x+x-\dfrac{1}{7}\right]\left[x^2+\dfrac{1}{49}-\dfrac{2}{7}x-x+\dfrac{1}{7}\right]=0\)
\(\Leftrightarrow\left(x^2+\dfrac{5}{7}x-\dfrac{6}{49}\right)\left(x^2-\dfrac{9}{7}x+\dfrac{8}{49}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+\dfrac{5}{7}x-\dfrac{6}{49}=0\\x^2-\dfrac{9}{7}x+\dfrac{8}{49}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{8}{7}\end{matrix}\right.\)
Vậy...
b, \(\left|x+6,4\right|+\left|x+2,5\right|+\left|x+8,1\right|=4x\)
\(\Leftrightarrow x+6,4+x+2,5+x+8,1=4x\) với mọi x
\(\Leftrightarrow x+x+x-4x=-8,1-2,5-6,4\)
\(\Leftrightarrow-x=-17\)
\(\Leftrightarrow x=17\)
Vậy...
câu a gồm : A(6: -2) , E( 0; 0)
câu b gồm : B( -2; -10 ) ,E ( 0: 0)
Bài 1:
a: \(=\dfrac{15-32}{40}\cdot10+\dfrac{1}{4}\)
\(=\dfrac{-17}{4}+\dfrac{1}{4}=-\dfrac{16}{4}=-4\)
b: \(=\left(\dfrac{9}{6}-\dfrac{5}{6}\right)^2+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{4}{9}+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{8}{18}+\dfrac{45}{18}+\dfrac{12}{18}=\dfrac{65}{18}\)
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