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a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)
Lời giải:
a)
PT \(\Leftrightarrow \frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Dễ thấy \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}<0\) nên $x+100=0$
$\Rightarrow x=-100$
b)
PT \(\Leftrightarrow \frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-73}{16}-1\)
\(\Leftrightarrow \frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow (x-89)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Dễ thấy \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}< 0\)
\(\Rightarrow x-89=0\Rightarrow x=89\)
a) Ta có: 3x(x-1)=(x-1)(x+2)
⇔3x(x-1)-(x-1)(x+2)=0
⇔(x-1)(3x-x-2)=0
⇔(x-1)(2x-1)=0
⇔2(x-1)2=0
mà 2≠0
nên (x-1)2=0
⇔x-1=0
hay x=1
Vậy: x=1
b) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)
\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)
\(\Leftrightarrow-181x-362=0\)
\(\Leftrightarrow-181x=362\)
hay x=-2
Vậy: x=-2
c) Ta có: \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+2\right)\)
\(\Leftrightarrow\frac{x}{2}+\frac{1}{2}+\frac{x}{4}+\frac{3}{4}=3-\frac{x}{2}-1\)
\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{5}{4}-\frac{-x}{2}-2=0\)
\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2}-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{x}{4}+x-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{x}{4}+\frac{4x}{4}-\frac{3}{4}=0\)
\(\Leftrightarrow5x-3=0\)
\(\Leftrightarrow5x=3\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
d) Ta có: \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
mà \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
nên x+100=0
hay x=-100
Vậy: x=-100
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left[\frac{x-12}{77}-1\right]+\left[\frac{x-11}{78}-1\right]=\left[\frac{x-74}{15}-1\right]-\left[\frac{x-73}{16}-1\right]\)
\(\Leftrightarrow\frac{x-12-77}{77}+\frac{x-11-78}{78}=\frac{x-74-15}{15}+\frac{x-73-16}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}=0\)
\(\Leftrightarrow\left[x-89\right]\cdot\left[\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right]=0\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
chị mk mới bày
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\frac{x-12-77}{77}+\frac{x-11-78}{78}=\frac{x-74-15}{15}+\frac{x-73-16}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=89\)
bạn sai đề hả? mình nghĩ là \(\frac{x-74}{75}+\frac{x-73}{76}\) câu này mình có làm rồi
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=0-100\)
\(\Leftrightarrow x=-100.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=0+89\)
\(\Leftrightarrow x=89.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)
Chúc bạn học tốt!
Câu g) bạn cộng 1 vào mỗi hạng tử của 2 vế
Câu h) bạn trừ một vào mỗi hạng tử ở hai vế
Quy đồng mẫu thì được tử giống nhau sau đó đặt nhân tử chung là xong