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\(\dfrac{x+1}{1}+\dfrac{2x+3}{3}+\dfrac{3x+5}{5}+...+\dfrac{10x+19}{19}=12+\dfrac{4}{3}+\dfrac{6}{5}+...+\dfrac{20}{19}\)
\(x+1+\dfrac{2x}{3}+1+\dfrac{3x}{5}+1+...+\dfrac{10x}{19}+1-12-\dfrac{4}{3}-\dfrac{6}{5}-...-\dfrac{20}{19}=0\)
\(x+\dfrac{2x}{3}-\dfrac{4}{3}+\dfrac{3x}{5}-\dfrac{6}{5}+...+\dfrac{10x}{19}-\dfrac{20}{19}+10-12=0\)
\(x-2+\dfrac{2x-4}{3}+\dfrac{3x-6}{5}+...+\dfrac{10x-20}{19}=0\)
\(x-2+\dfrac{2\left(x-2\right)}{3}+\dfrac{3\left(x-2\right)}{5}+...+\dfrac{10\left(x-2\right)}{19}=0\)
\(\left(x-2\right)\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)=0\)
Ta thấy \(\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)>0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
tìm x
a/\(\dfrac{x}{2}=\dfrac{32}{x}\)
b/\(\dfrac{5}{25}=5\)
c/\(\dfrac{30^x}{27^x}=27\)
d/5x . 10x=2500
Đặt x/3=y/5=k
=>x=3k; y=5k
\(A=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{5\cdot9+3\cdot25}{10\cdot9-3\cdot25}=8\)
\(f\left(x\right)=10x+5\)
Nghiệm của \(f\left(x\right)\) là \(x\Leftrightarrow10x+5=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\) (1)
\(g\left(x\right)=x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}\)
Nghiệm của \(g\left(x\right)\)là \(x\Leftrightarrow x^3+\dfrac{1}{2}x^2+3x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x^2\left(x+\dfrac{1}{2}\right)+3\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x^2+3=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\) (2)
Từ (1) và (2) => ĐPCM
\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\\ \Leftrightarrow\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}\)
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)
=>\(4x+6=\dfrac{1}{2}\left(10x+4\right)\)
=>4x+6=5x+2
=>x=6-2=4
Vậy x=4
Mình sẽ làm cách dãy tỉ số bằng nhau ,vì nhân sẽ khá là rối =.=
\(\dfrac{2x+3}{5x+2}=\dfrac{2\left(2x+3\right)}{2\left(5x+2\right)}=\dfrac{4x+6}{10x+4}\)
Hay \(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)
Thay vào ta có:
\(\dfrac{2x+3}{5x+2}=\dfrac{1}{2}\Leftrightarrow5x+2=4x+6\Leftrightarrow5x=4x+4\Leftrightarrow x=4\)
Giải:
Ta có: \(\dfrac{3x-2y}{5}=\dfrac{5y-3z}{2}=\dfrac{2z-5x}{2}\)
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}=\dfrac{15x-10y+10y-6z+6z-15x}{25+4+6}=0\)
\(\Rightarrow\left\{{}\begin{matrix}15x-10y=0\\10y-6z=0\\6z-15x=0\end{matrix}\right.\Rightarrow15x=10y=6z\)
\(\Rightarrow\dfrac{15x}{30}=\dfrac{10y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=10\\y=15\\z=25\end{matrix}\right.\)
Vậy...
\(\dfrac{3x-2y}{5}=\dfrac{5y-3z}{2}=\dfrac{2z-5x}{2}\)
\(\Rightarrow\dfrac{5\left(3x-2y\right)}{25}=\dfrac{2\left(5y-3z\right)}{4}=\dfrac{3\left(2z-5x\right)}{6}\)
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
\(=\dfrac{15x-10y+10y-6z+6z-15x}{25+4+6}\)
\(=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\5y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{5}\\2z=5x\Rightarrow\dfrac{z}{5}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.5=25\end{matrix}\right.\)
Ta có :
\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\)
\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=4x\left(5x+2\right)+5\left(5x+2\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+8x+25x+10\)
\(\Leftrightarrow20x^2+34x+6=20x^2+33x+10\)
\(\Leftrightarrow20x^2+34x+6-20x^2-33x-10=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Vậy x = 4
D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24
thiếu 1 câu
A= x5−5x4+5x3−5x2+5x−1x5−5x4+5x3−5x2+5x−1 với x = 4
= x5−(x+1)x4+(x+1)x3−(x+1)x2+(x+1)x−1
= x5−x5−x4+x4+x3−x3+x2−x2+x−1
=x−1=4−1=3
Tương tự với các câu B,C,D
\(5^x+10x=625+10x\)
\(\Rightarrow5^x=625+10x-10x\)
\(\Rightarrow5^x=625\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\)
Vậy x=4