\(\frac{x}{4}\)=\(\frac{15}{20}\)
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a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
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\(\frac{x+5}{95}+\frac{x+10}{90}+\frac{x+15}{85}+\frac{x+20}{80}=-4\)
<=> \(\frac{x+5}{95}+1+\frac{x+10}{90}+1+\frac{x+15}{85}+1+\frac{x+20}{80}+1=0\)
<=> \(\frac{x+100}{95}+\frac{x+100}{90}+\frac{x+100}{85}+\frac{x+100}{80}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{90}+\frac{1}{85}+\frac{1}{80}\right)=0\)
<=> \(x+100=0\) (do 1/95 + 1/90 + 1/85 + 1/80 khác 0)
<=> \(x=-100\)
Vậy...
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\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
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\(e,\frac{22}{15}-x=-\frac{8}{27}\)
=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)
=> \(x=\frac{22}{15}+\frac{8}{27}\)
=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)
\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)
=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)
=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)
=> \(\frac{2x-5}{5}=-\frac{5}{4}\)
=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)
=> \(2x=-\frac{5}{4}\)
=> \(x=-\frac{5}{8}\)
\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)
=> \(-\frac{9}{4}x+\frac{37}{4}=20\)
=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)
=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)
\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)
=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)
=> \(-10\le x\le-\frac{13}{7}\)
Đến đây tìm x
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a, \(x:3\frac{1}{15}=1\frac{1}{2}\)
\(\Rightarrow x=1\frac{1}{2}\cdot3\frac{1}{15}\)
\(\Rightarrow x=\frac{3}{2}\cdot\frac{46}{15}=\frac{3\cdot46}{2\cdot15}=\frac{1\cdot23}{1\cdot5}=\frac{23}{5}=4\frac{3}{5}\)
\(b)x\cdot\frac{15}{28}=\frac{3}{20}\)
\(\Rightarrow x=\frac{3}{20}:\frac{15}{28}=\frac{3}{20}\cdot\frac{28}{15}=\frac{1}{5}\cdot\frac{7}{5}=\frac{7}{25}\)
Tự làm nốt câu cuối :>
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cộng 1 vào từ số hạng ( hai vế cùng 3 số hạng=> không đổi)
Tử số còn lại x
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)
\(\Leftrightarrow\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)x=0\)
cái (...) khác không=> x =0 là nghiệm duy nhất
Ta có
\(\frac{x-19}{19}+\frac{x-20}{20}+\frac{x-21}{21}=\frac{x-17}{17}+\frac{x-16}{16}+\frac{x-15}{15}\)
\(\Leftrightarrow\frac{x}{19}-1+\frac{x}{20}-1+\frac{x}{21}-1=\frac{x}{17}-1+\frac{x}{16}-1+\frac{x}{15}-1\)
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-3=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}-3\)
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)
\(\Rightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-\frac{x}{17}-\frac{x}{16}-\frac{x}{15}=0\)
\(\Leftrightarrow x\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)
Vì \(\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)\ne0\)
Nên phương trình chỉ co nghiệm duy nhất là x=0
Vậy x=0
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1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)
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a) Ta có:
\(\begin{array}{l}\frac{x}{3} = \frac{y}{4} \Rightarrow \frac{x}{3}.\frac{1}{5} = \frac{y}{4}.\frac{1}{5} \Rightarrow \frac{x}{{15}} = \frac{y}{{20}};\\\frac{y}{5} = \frac{z}{6} \Rightarrow \frac{y}{5}.\frac{1}{4} = \frac{z}{6}.\frac{1}{4} \Rightarrow \frac{y}{{20}} = \frac{z}{{24}}\end{array}\)
Vậy \(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}}\) (đpcm)
b) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}} = \frac{{x - y + z}}{{15 - 20 + 24}} = \frac{{ - 76}}{{19}} = - 4\)
Vậy x = 15 . (-4) = -60; y = 20. (-4) = -80; z = 24 . (-4) = -96
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tìm n N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên
giúp mik với sắp thi r
\(\frac{x}{4}=\frac{15}{20}\)
\(x:4=\frac{15}{20}\)
\(x=\frac{15}{20}\times4\)
\(x=3\)
Vậy \(x=3\)
x = 4 x 15 :20 = 3