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a, \(2x-\frac{1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{1}{2}\left(4x-1\right)=\frac{1}{8}\left(6x+1\right)\)
\(\Leftrightarrow4\left(4x-1\right)=6x+1\)
\(\Leftrightarrow10x=5\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b, \(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x = 3
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=0+3\)
\(\Leftrightarrow x=3\)
Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.
Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x - 100 = 0
=> x = 100
Gợi ý :
Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
Pro kinh không thể lướt qua
Giao luu: x=3
có thể quy đồng làm bt; \(!vt!\ge!vp!\)
Hay dùng bpt vào giải pt