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\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4031}{2015^2.2016^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{2016^2-2015^2}{2015^2.2016^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
\(A=1-\dfrac{1}{2016^2}< 1\left(đpcm\right)\)
\(\dfrac{72-x}{7}=\dfrac{x-4}{9}\)
\(\Rightarrow9\left(72-x\right)=7\left(x-4\right)\)
\(\Rightarrow648-9x=2x-28\)
\(\Rightarrow11x-28=648\)
\(\Rightarrow11x=676\Rightarrow x=\dfrac{676}{11}\)
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Rightarrow259-7x=3x+39\)
\(\Rightarrow10x+39=259\)
\(\Rightarrow10x=220\Rightarrow x=22\)
\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=\pm10^2\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\Rightarrow x=6\\x+4=-10\Rightarrow x=-14\end{matrix}\right.\)
\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\)
\(\Rightarrow x\left(x+3\right)-1\left(x+3\right)=x\left(x+2\right)-2\left(x+2\right)\)
\(\Rightarrow x^2+3x-x-3=x^2+2x-2x-4\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x-3=-4\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\dfrac{1}{2}\)
a) Ta có:
+) a/2=b/3
=>a=2b/3
+) b/5=c/4
=>c=4b/5
Lại có:
a-b+c=49
=> 2b/3 -b + 4b/5 =49
=> 7b/15==49
=> b= 105
Khi đó:
+) a=2b/3=2.105/3=70
+)c=4b/5=4.105/5=84
Vậy a=70; b=105; c=84...
chúc bạn học tốt
a,|x2−13x2−13| = 3232
b, 32−1232−12 ( 2x-1)=3434
c, |x-1|+2x=2
a)\(\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{3}{2}\)
TH1
\(\dfrac{x}{2}-\dfrac{1}{3}=\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=\dfrac{11}{6}\)
=>x=\(\dfrac{11.2}{6}\)
=>x=\(\dfrac{11}{3}\)
TH2
\(\dfrac{x}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=-\dfrac{3}{2}+\dfrac{1}{2}\)
=>\(\dfrac{x}{2}=-1\)
=>x=-2
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)
\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
2)
a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72
\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)
\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)
Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)
Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)
b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12
\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)
\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)
Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)
Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)
cho hỏi chút
\(\frac{a}{b}=\frac{c}{d}\)
trong đó
\(a=c\) hay \(a\ne c\)
\(b=d\) hay \(b\ne d\)
( bài có thiếu điều kiện ko vậy )
\(\dfrac{-4}{13}\cdot\dfrac{5}{17}+\dfrac{-12}{13}\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{-4}{17}\cdot\dfrac{5}{17}+\dfrac{-4}{13}\cdot3\cdot\dfrac{4}{17}+\dfrac{4}{13}\)
\(=\dfrac{4}{13}\left(\dfrac{-5}{17}+\dfrac{-12}{17}+1\right)\)
\(=\dfrac{4}{13}\left(-1+1\right)=\dfrac{4}{13}\cdot0=0\)
Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}\)
\(\Rightarrow2A=\frac{2}{2^3}+\frac{2}{3^3}+...+\frac{2}{2019^3}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{2}{2^3}< \frac{2}{1.2.3}\\\frac{2}{3^3}< \frac{1}{2.3.4}\\....\\\frac{2}{2019^3}< \frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\end{matrix}\right.\)
\(\Rightarrow2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(2019-1\right).2019}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.2020}\)
\(\Rightarrow A< \left(\frac{1}{1.2}-\frac{1}{4078380}\right):2\)
\(\Rightarrow A< \frac{1}{1.2}:2-\frac{1}{4078380}:2\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{8156760}\)
\(\Rightarrow A< \frac{1}{2^2}-\frac{1}{8156760}\)
Vì \(\frac{1}{2^2}-\frac{1}{8156760}< \frac{1}{2^2}.\)
\(\Rightarrow A< \frac{1}{2^2}\left(đpcm\right).\)
Chúc bạn học tốt!