Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Rightarrow\frac{x-12}{77}-1+\frac{x-11}{78}-1=\frac{x-74}{15}-1+\frac{x-73}{16}-1\)
\(\Rightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\Rightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}\right)=\left(x-89\right).\left(\frac{1}{15}+\frac{1}{16}\right)\)
=> \(\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}\right)-\left(x-89\right)\left(\frac{1}{15}+\frac{1}{16}\right)=0\)
=> \(\left(x-89\right).\left[\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)\right]=0\Rightarrow x-89=0\left(\text{vì }\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0\right)\)
=> x = 89
Vậy x = 89
\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-12}{77}+\dfrac{x-11}{78}-\dfrac{x-74}{15}-\dfrac{x-73}{16}=0\)
\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1-\dfrac{x-74}{15}+1-\dfrac{x-73}{16}+1=0+1+1-1-1\)
\(\Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)-\left(\dfrac{x-74}{15}-1\right)-\left(\dfrac{x-73}{16}-1\right)=0\)
\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-89=0\\\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\end{matrix}\right.\)
\(x-89=0\\ \Rightarrow x=89\)
\(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\)(vô lí)
Vậy \(x=89\)
easy làm câu b vs c trước nha
b) \(\left(x-5\right)\left(2x+4\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\2x+4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\2x+4< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\)
Vậy......
c) \(\left(x+3\right)\left(3x-6\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\3x-6< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\3x-6>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3< x< 2\\x\in\varnothing\end{matrix}\right.\)
Vậy.......
71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 + 79
= ( 71 + 79 ) + ( 72 + 78 ) + ( 73 + 77 ) + ( 74 + 76 ) + 75
= 150 + 150 + 150 + 150 + 75
= 150 x 4 + 75
= 600 + 75
= 675
Tìm x :
X x 7 + X x 2 = 108
X x ( 7 + 2 ) = 108
X x 9 = 108
X = 108 : 9
X = 12
\(71+72+73+74+75+76+77+78+79\)
\(=\left(71+79\right)+\left(72+78\right)+\left(73+77\right)+\left(74+76\right)+75\)
\(=150+150+150+150+75\)
\(=600+75\)
\(=675\)
\(X\)x\(7+X\)x\(2=108\)
\(X\)x\(\left(7+2\right)=108\)
\(X\)x\(9=108\)
\(X=108:9\)
\(X=12\)
Vậy \(X=12\)
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
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)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
<=> \(\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
<=> \(\frac{x-12-77}{77}+\frac{x-11-78}{78}=\frac{x-74-15}{15}+\frac{x-73-16}{16}\)
<=> \(\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
<=> \(\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
<=> \(\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
<=> x - 89 = 0 <=> x = 89
Các Admin ơi hiện nay có một bạn tên là Quản lý Online Math nhưng đây không phải là quản lí mà là Nam Cao Nguyễn bạn ấy thương xuyên bảo chúng mình đặt bảo mật rôi bây giờ cậu ấy lấy nick của Nguyễn Thị Hiện Nhân
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 ) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}+\dfrac{x+4}{97}=-4\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}+\dfrac{x+101}{97}=-1\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{97}\right)=-1\)
Vô lí => Phương trình trên vô nghiệm .
b ) \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-73}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89.
Ta có:
x − 2 77 + x − 1 78 = x − 74 5 + x − 73 6 ⇔ x − 2 77 − 1 + x − 1 78 − 1 = x − 74 5 − 1 + x − 73 6 − 1 ⇔ x − 2 − 77 77 + x − 1 − 78 78 = x − 74 − 5 5 + x − 73 − 6 6 ⇔ x − 79 77 + x − 79 78 = x − 79 5 + x − 79 6 ⇔ x − 79 1 77 + 1 78 − 1 5 − 1 6 = 0
Nhận thấy 1 77 + 1 78 − 1 5 − 1 6 ≠ 0
Nên x - 79 = 0 ⇔ x = 79
Vậy x = 79 là nghiệm của phương trình.
Đáp án cần chọn là: A