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1. đề bạn ghi rõ lại giúp mình đc ko r mình giải lại cho
2. Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{y^2}{5^2}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)
\(\dfrac{x}{3}=4\Rightarrow x=12\)
\(\dfrac{y}{5}=4\Rightarrow y=20\)
Vậy x=12 và y=20
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
=>-6a+5b=6a-5b
=>-12a=-10b
=>6a=5b
hay a/b=5/6
Bài 1:
a: \(x=-\dfrac{7}{11}=-1+\dfrac{4}{11}\)
\(y=-\dfrac{13}{17}=-1+\dfrac{4}{17}\)
mà 4/11>4/17
nên x>y
b: \(x=\dfrac{987}{863}=1+\dfrac{124}{863}\)
\(y=\dfrac{398}{274}=1+\dfrac{124}{274}\)
mà 124/863<124/274
nên x<y
Để \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) thì a(b+d) < b(a+c)
<=> ab + ad < ba + cb
<=> ad < cb
<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)
Để \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) thì (a+c)d < (b+d)c
<=> ad + cd < bc + dc
<=> ad < bc
<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)
Chúc bạn học tốt!
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)
\(\Rightarrow\) \(x=\dfrac{1}{5}\)
a. \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Leftrightarrow\dfrac{1}{15}:\left(2x\right)=0,75:4,5\)
\(\Rightarrow\dfrac{1}{15}:\left(2x\right)=\dfrac{1}{6}\)
\(\Rightarrow2x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{2}{5}:2=\dfrac{1}{5}\)
Vậy...
b. \(\dfrac{-5}{x-2}=\dfrac{3}{-9}\)
\(\Leftrightarrow\left(x-2\right).3=\left(-5\right).\left(-9\right)\)
\(\Rightarrow\left(x-2\right).3=45\)
\(\Rightarrow\left(x-2\right)=45:3=15\)
\(\Rightarrow x=15+2=17\)
Vậy...
c. \(\dfrac{-2}{3}:x=\dfrac{1}{2}:\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-2}{3}:x=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}:\dfrac{2}{3}=-1\)
Vậy...
Với \(a,b,c\ne0\) ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=1\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
Vì \(a=b=c\Rightarrow\dfrac{a^{49}\times b^{51}}{c^{100}}=\dfrac{a^{49}\times a^{51}}{a^{100}}=\dfrac{a^{100}}{a^{100}}=1\)
Chúc bn học tốt 
(Bài này phải đc gọi là "Ác mộng dấu bằng")