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=> \(\left(x-3\right).\left(x-3\right)=5.\left(9.8\right)\)
=> \(\left(x-3\right)^2=49\)
=> \(\orbr{\begin{cases}\left(x-3\right)^2=7^2\\\left(x-3\right)^2=\left(-7\right)^2\end{cases}}\) => \(\orbr{\begin{cases}x-3=7\\x-3=-7\end{cases}}\)=> \(\orbr{\begin{cases}x=7+3\\x=-7+3\end{cases}}\)=> \(\orbr{\begin{cases}x=10\\x=-4\end{cases}}\)
Vậy : x \(\varepsilon\){ 10 ; -4 }
P/s : \(\orbr{\begin{cases}\\\end{cases}}\) nghĩa là hoặc
(x-3)2=5.9,8
(x-3)2=49
TH1: x-3=7
=>x=7+3=10
TH2:x-3=-3
=> x=-3+3=0
Vay x=10;0
![](https://rs.olm.vn/images/avt/0.png?1311)
Dài đấy :))
a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)
\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)
\(\Leftrightarrow\left|x-1\right|+8=9\)
\(\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)
c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))
\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)
\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)
\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)
d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)
\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)
\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)
\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)
Vậy ta xét hai trường hợp sau :
1. \(x\ge-\frac{3}{16}\)
(*) <=>\(7x-2=4x+\frac{3}{4}\)
\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)
\(\Leftrightarrow3x=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)
2. \(x< -\frac{3}{16}\)
(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)
\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)
\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)
\(\Leftrightarrow11x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)
Vậy x = 11/12
e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)
\(\Leftrightarrow x+1=4040\)
\(\Leftrightarrow x=4039\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, <=> 7(37-x)=3(x+13)
<=> x =22
b, <=> (x+1)(x-1)=15
<=> x^2-1=15 <=> x^2=16 <=> x= +_4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{4}x-\frac{2}{3}.\left(\frac{3}{5}x-\frac{6}{5}\right)=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{3}{4}x-\frac{2}{5}x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\left(\frac{3}{4}-\frac{2}{5}\right)x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\left(\frac{15}{20}-\frac{8}{20}\right)x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{7}{20}x+\frac{4}{5}=\frac{1}{7}-\frac{2}{9}x\)
\(\frac{1}{7}-\frac{4}{5}=\frac{2}{9}x-\frac{7}{20}x\)
\(\frac{5}{35}-\frac{28}{35}=\left(\frac{2}{9}-\frac{7}{20}\right)x\)
\(\frac{-23}{35}=\left(\frac{40}{180}-\frac{63}{180}\right)x\)
\(\frac{-23}{180}x=\frac{-23}{35}\)
\(x=\frac{-23}{35}:\frac{-23}{180}\)
\(x=\frac{-23}{35}.\frac{180}{-23}\)
\(x=\frac{180}{35}\)
Vậy \(x=\frac{180}{35}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x+\frac{1}{5}\right|-4=-2\)
=) \(\left|x+\frac{1}{5}\right|=-2+4=2\)
=) \(x+\frac{1}{5}=2\)hoặc \(x+\frac{1}{5}=-2\)
=) \(x=2-\frac{1}{5}=\frac{9}{5}\); =) \(x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\)
Vậy \(x=\left\{\frac{9}{5},\frac{-11}{5}\right\}\)
b)\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
=) \(2x-\frac{6}{5}x=\frac{-1}{2}+\frac{1}{5}\)
=) \(x.\left(2-\frac{6}{5}\right)=\frac{-3}{10}\)
=) \(x.\frac{4}{5}=\frac{-3}{10}\)
=) \(x=\frac{-3}{10}:\frac{4}{5}\)
=) \(x=\frac{-3}{8}\)
c) \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
=) \(\left(x-3\right)^{x+2}.\left(1-6\right)=0\)
=) \(\left(x-3\right)^{x+2}=0:\left(1-6\right)=0\)
Mà chỉ có \(0^x=0\)
=) \(x-3=0\)
=) \(x=0+3\)
=) \(x=3\)
a,
\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}\)
b,
\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\Leftrightarrow x=-\frac{3}{8}\)
c,
\(\left[x-3\right]^{x+2}-\left[x-3\right]^{x+8}=0\)
=> [x-3]x + 2 = [x-3]x+8
=> x + 2 = x + 8
=> x không tồn tại
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x-1/65-1-x-3/63-1+x-5/61-1+x-7/59-1 x-66/65-x-66/63+x-66/61+x-66/59 =0 suy ra (x-66).(1/65-1/63+1/61+1/59)=0 vi 1/65-1/63+1/61+1/59khong thuoc 0 nen x-66+66=0 suy ra x =132
a-b=2.(a+b) tương đương a-b =2a + 2b tương đương -3b=a
a-b=a.b suy ra -3b-b=-3b.b tương đương -4b=-3b.b tương đương b=4/3 suy ra a=-4
với a=-4 ; b=4/3 thì a-b = 2.(a+b)= a.b
![](https://rs.olm.vn/images/avt/0.png?1311)
he he !! linh học giỏi hơn mình lun !! mình học dở ẹc àk !!
Ta có:
\(\frac{x+3}{3}=\frac{27}{x-3}\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=27.3\)
\(\Leftrightarrow x^2-9=27.3\)
\(\Leftrightarrow x^2-9=81\)
\(\Leftrightarrow x^2=81+9\)
\(\Leftrightarrow x^2=90\)
\(\Leftrightarrow x=\sqrt{90}=3\sqrt{10}\)
Cái đoạn(x+3)(x-3)=x2-9 là mình dùng hằng đẳng thức của lớp 8
x+3/3=27/x-3
=> (x+3).(x-3)=27.3
=> x.(x-3)+3.(x-3)=81
=> x2-3x+3x-9=81
=> x2-9=81
=> x2=72
=> x= căn 72