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a) \(\left(\frac{3}{5}-1\frac{3}{4}\right)\div2\frac{3}{10}=\left(\frac{3}{5}-\frac{7}{4}\right)\div\frac{23}{10}=\frac{-23}{20}.\frac{10}{23}=-\frac{1}{2}\)
b) \(\left(\frac{-2}{3}+\frac{3}{4}\right)^2.\frac{12}{5}-\frac{11}{5}=\left(\frac{1}{12}\right)^2.\frac{12}{5}-\frac{11}{5}=\frac{1}{60}-\frac{11}{5}=-\frac{33}{15}\)
c) \(\frac{-7}{8}\div\frac{21}{16}-\frac{5}{3}\left(\frac{1}{3}-\frac{7}{10}\right)=\frac{-7}{8}\times\frac{16}{21}-\frac{5}{3}.-\frac{11}{30}=-\frac{2}{3}-\frac{11}{18}=-\frac{23}{18}\)
\(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)
\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)\)
\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\). Vì \(\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)>0\)
\(\Leftrightarrow x-14=0\Rightarrow x=0+14=14\). Vậy \(x=14\)
Bài làm:
Ta có: \(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)
\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)=0\)
\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\)
\(\Leftrightarrow x-14=0\)
\(\Rightarrow x=14\)
Ta có: \(\frac{x}{3}=\frac{y}{7}\)\(\Rightarrow\)\(x=\frac{3y}{7}\)
Ta lại có: \(\frac{2}{x}=\frac{y}{-3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow\frac{3y}{7}.y=-6\)
\(\Leftrightarrow y^2=-6.\frac{7}{3}\)
\(\Leftrightarrow y^2=-14\)
Vì \(\hept{\begin{cases}y^2\ge0\forall y\\-14< 0\end{cases}}\)mà \(y^2=-14\)
\(\Rightarrow\)\(y\in\varnothing\)\(\Rightarrow\)\(x\in\varnothing\)
Vậy x,y không có giá trị
Bài 2 :
a, \(P\left(x\right)=2x^5+2-6x^2-3x^3+4x^2-2x+x^3+4x^5=6x^5-2x^3-2x^2+2\)
b, sắp xếp rồi, trên ý
c, Bậc : 5
Bài 3 : \(Q\left(x\right)=3x-5=0\Leftrightarrow x=\frac{5}{3}\)
Vì\(\frac{121212}{242424}=\frac{12.20202}{24.20202}=\frac{1}{2}\)nên\(\frac{121212}{242424}=\frac{1}{2}\).
Ta có : \(\frac{121212}{242424}\)= \(\frac{1.121212}{2.121212}\)=\(\frac{1}{2}\)
2009 - | x - 2009 | = x
=>| x - 2009 | = 2009 - x
=> x = 2009
\(|x-2009|=2009-x\) ( 1 )
\(ĐK:2009-x\ge0\)
\(-x\ge0-2009\)
\(-x\ge-2009\)
\(x\le2009\)
( 1 ) \(\Leftrightarrow\orbr{\begin{cases}x-2009=2009-x\\x-2009=-\left(2009-x\right)\end{cases}}\)
\(\orbr{\begin{cases}x+x=2009+2009\\x-2009=-2009+x\end{cases}}\)
\(\orbr{\begin{cases}2x=4018\\x-x=-2009+2009\end{cases}}\)
\(\orbr{\begin{cases}x=2009\\0x=0\left(llđ\forall x\right)\end{cases}}\)
Vậy \(x\le2009\) là nghiệm của phương trình
Gỉa sử : \(\frac{a}{b}< \frac{a+c}{b+c}< =>ab+ac< ab+bc\)
\(< =>ac< bc< =>a< b\)(đpcm)
Gỉa sử : \(\frac{a}{b}>\frac{a+c}{b+c}< =>ab+ac>ab+bc\)
\(< =>ac>bc< =>a>b\)(đpcm)
a) Rút gọn :
P = l 3x - 3 l + 2x + 1 ( Vì 3x ≥ 3 => x ≥ 3 ; 3x - 3 ≥ 0 nên l 3x -3 l = 3x -3 )
=> 3x - 3 + 2x + 1 = 5x - 2
b) Tính giá trị x để P = 6
<=> 5x -2 = 6
<=> 5x = 8
x = \(\frac{8}{5}\)
a) \(K=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{2^8.3^5}=\frac{2^3}{3}=\frac{8}{3}\)
b) \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{3^6.3^6}{3^2.2^2.3^{10}}=\frac{1}{2^2}=\frac{1}{4}\)
c) \(P=\frac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\frac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\frac{2.3}{5}=\frac{6}{5}\)