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a/
Đặt $\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}=k$
$\Rightarrow a=2k+1; b=3k+2; c=4k+3$
Khi đó:
$3a+3b-c=50$
$\Rightarrow 3(2k+1)+3(3k+2)-(4k+3)=50$
$\Rightarrow 11k+6=50$
$\Rightarrow 11k=44\Rightarrow k=4$
Ta có:
$a=2k+1=2.4+1=9$
$b=3k+2=3.4+2=14$
$c=4k+3=4.4+3=19$
b/
$2a=3b; 5b=7c\Rightarrow \frac{a}{3}=\frac{b}{2}; \frac{b}{7}=\frac{c}{5}$
$\Rightarrow \frac{a}{21}=\frac{b}{14}=\frac{c}{10}$
Áp dụng TCDTSBN:
$\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{45}{15}=3$
$\Rightarrow a=21.3=63; b=14.3=42; c=10.3=30$
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
\(b,\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+\frac{2}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},-\frac{2}{3}\right\}\)
a) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-\frac{1}{6}-5x-\frac{3}{25}=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-5x+x=\frac{1}{5}+\frac{1}{6}+\frac{3}{25}\)
\(\Leftrightarrow-x=\frac{73}{150}\)
\(\Leftrightarrow x=-\frac{73}{150}\)
Vậy : \(x=-\frac{73}{150}\)
áp dụng dãy tỉ số = nhau ta có
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a-b}{c-d}=\frac{a+b}{c-d}\)
Ta xét
Vế 1 \(\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{ab}{cd}\)( nhân cả tử mẫu lại với nhau )
Vế 2 : \(\frac{a-b}{c-d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a-b\right)\left(a+b\right)}{\left(c-d\right)\left(c+d\right)}=\frac{a^2-b^2}{c^2-d^2}\) ( nhân cả tử cả mẫu với nhau )
Mà Vế 1 = vế 2
=> \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\)
\(\frac{x-1}{4}=\frac{2x+1}{5}\)
\(\Rightarrow5\left(x-1\right)=4\left(2x+1\right)\)
\(\Rightarrow5x-5=8x+4\)
\(\Rightarrow5x-8x=4+5\)
\(\Rightarrow-3x=9\)
\(\Rightarrow x=-3\)
vậy_
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+x+2x-x^2+3x+x=3-2\)
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
vậy_
Ta có : C = y . \(\frac{8}{5}.x.ab^5.2.x^3.y\)
= \(\frac{16}{5}.a.b^5.x^4.y^2\)
Trong đó : hệ số : \(\frac{16}{5}.a.b^5\)
: biến : x ; y
: bậc : 4,2
\(\frac{3a^2-b^2}{a^2+b^2}=\frac{3}{4}\)
<=> \(4\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
<=> \(12a^2-4b^2=3a^2+3b^2\)
<=> \(9a^2=7b^2\)
<=> \(\frac{a^2}{b^2}=\frac{7}{9}\)
<=> \(\frac{a}{b}=\pm\frac{\sqrt{7}}{3}\)