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a) Ta có:
\(\frac{x}{x+4}=\frac{5}{6}\)
\(\Rightarrow6x=5\left(x+4\right)\)
\(\Rightarrow6x=5x+20\)
\(\Rightarrow5x+20-6x=0\)
\(-x=-20\)
\(x=20\)
b)Ta có: \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-21-5x-25=0\)
\(2x=46\)
\(x=23\)
c)Ta có: \(\frac{x+4}{20}=\frac{5}{20}\)
\(\Rightarrow x+4=5\)
\(\Rightarrow x=1\)
a) \(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{2}{5}+\frac{5}{7}\)
\(x=\frac{14}{35}+\frac{25}{35}=\frac{39}{35}\)
b)
\(\frac{-2}{5}x=\frac{4}{15}\)
\(x=\frac{4}{15}:-\frac{2}{5}\)
\(x=\frac{4}{15}\cdot-\frac{5}{2}=-\frac{2}{3}\)
c) \(2x\left(x-\frac{1}{7}\right)=2x^2-\frac{2x}{7}\)
d) \(\frac{1}{2}+\frac{3}{4}x=\frac{1}{4}\)
\(\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{3}{4}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}\cdot\frac{4}{3}=-\frac{1}{3}\)
f) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{5}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{5}=\frac{31}{60}\)
\(x=\frac{31}{60}-\frac{2}{5}=\frac{7}{60}\)
mk làm mẫu 2 bài đầu nhé, các bài còn lại bạn làm tương tự, các bài này đều áp dụng tính chất dãy tỉ số bằng nhau
1) Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{14}{7}=2\)
suy ra: \(\frac{x}{3}=2\)=> \(x=6\)
\(\frac{y}{4}=2\)=> \(y=8\)
Vậy...
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{3}=\frac{x-y}{5-3}=\frac{20}{2}=10\)
suy ra: \(\frac{x}{5}=10\)=> \(x=50\)
\(\frac{y}{3}=10\)=> \(y=30\)
Vậy...
a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy \(x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)
\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)
\(\Rightarrow x^2-x-x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10\)
+) \(x+4=10\Rightarrow x=6\)
+) \(x+4=-10\Rightarrow x=-16\)
Vậy \(x\in\left\{6;-16\right\}\)
Ta có: \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-3=5x+25\)
\(\Rightarrow7x-5x=25+3\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
- c, Ta co :\(\frac{x+4}{20}=\frac{5}{x+4}\)
=> \(\left[x+4\right].\left[x+4\right]=20.5\)
=> \(\left[x+4\right]^2=100\)
=> \(\left[x+4\right]^2=10^2\)
=> \(x+4=10\)
=> \(x=10-4\)
=> \(x=6\)
k mk nha
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a) \(-\frac{3}{x}=\frac{15}{7}\)
=> -3.7 = 15x
=> 15x = -21
=> x = -21:15
=> x = -1,4
Vậy x = -1,4
b) \(\frac{x+3}{4}=\frac{5}{20}\)
\(\Rightarrow\frac{x+3}{4}=\frac{1}{4}\)
=> x + 3 = 1
=> x = 1 - 3
=> x = -2
Vậy x = -2
d) \(\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5(x - 1) = 3(x + 1)
=> 5x - 5 = 3x + 3
=> 5x - 3x = 5 + 3
=> 2x = 8
=> x = 8:2
=> x = 4
Vậy x = 4
\(a,\frac{-3}{x}=\frac{15}{7}\)
=> -21 = 15x
=> \(x=-\frac{21}{15}=-\frac{7}{5}\)
b,
\(\frac{x+3}{4}=\frac{5}{20}\)
=> \(\frac{5(x+3)}{20}=\frac{5}{20}\)
=> 5\((x+3)\)= 5
=> x + 3 = 1
=> x = -2
\(c,\frac{1,2}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{12}{10}}{30}=\frac{3x+4}{50}\)
=> \(\frac{\frac{6}{5}}{30}=\frac{3x+4}{50}\)
=> \(\frac{2}{50}=\frac{3x+4}{50}\)
=> 3x + 4 = 2
=> 3x = -2
=> x = -2/3
\(d,\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5[x - 1] = 3[x + 1]
=> 5x - 5 = 3x + 3
=> 5x - 5 - 3x = 3
=> 5x - 3x - 5 = 3
=> 2x = 8
=> x = 4
\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)
\(=>6x-5x=20=>x=20\)
áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)