giải phương trình:
x^2-10x-29/1971+x^2-10x-27/1973=x^2-10x-1971/29+x^2-10x-1973/27
các bn giúp mik vs mai mik phải nộp r
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 2 2021
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
9 tháng 2 2021
a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)
\(\Rightarrow x-100=0\Rightarrow x=100\)
Vậy x = 100
25 tháng 5 2021
`(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/1929+(x^2-10x-1973)/1927`
`<=>(x^2-10x-29)/1971-1+(x^2-10x-27)/1973-1=(x^2-10x-1971)/1929-1+(x^2-10x-1973)/1927-1`
`<=>(x^2-10x-200)/1971+(x^2-10x-200)/1973=(x^2-10x-200)/1971+(x^2-10x-200)/1927`
`<=>(x^2-10x-200)(1/1971+1/1973-1/1929-1/1927)=0`
`<=>x^2-10x-200=0` do `1/1971+1/1973-1/1929-1/1927<0`
`<=>x^2-20x+10x-200=0`
`<=>x(x-20)+10(x-20)=0`
`<=>(x-20)(x+10)=0`
`<=>` \(\left[ \begin{array}{l}x=20\\x=-10\end{array} \right.\)
Vậy `S={20,-10}`
B
17 tháng 4 2017
Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-27}{1973}-1\right)+\left(\dfrac{x^2-10x-29}{1971}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1973}+\dfrac{x^2-10x-2000}{1971}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)=0\) vì \(\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)\ne0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy: Giá trị x thỏa mãn là: \(x=-40;50\)
C
17 tháng 4 2017
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
vì \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
Nên \(x^2-10x-2000=0\)
<=> \(x^2-50x+40x-2000=0\)
<=> \(x\left(x-50\right)+40\left(x-50\right)=0\)
<=> \(\left(x-50\right)\left(x+40\right)=0\)
<=> \(x=50\) hoặc \(x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)
M
11 tháng 2 2016
\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1972}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\frac{x^2-10x-29-1971}{1971}+\frac{x^2-10x-27-1973}{1973}=\frac{x^2-10x-1971-29}{29}+\frac{x^2-10x-1973-27}{27}\)
\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x^2-10x-2000=0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
Th1: \(x-50=0\Leftrightarrow x=50\)
Th2: \(x+40=0\Leftrightarrow x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)
4 tháng 2 2017
Thêm (-1) vào từng số hạng=> tử số các số hạng là: \(\left(x^2-10x-2000\right)\)
\(\Leftrightarrow x^2-10x-2000=0\Leftrightarrow\left(x-5\right)^2=2025=45^2\)
\(\orbr{\begin{cases}x=50\\x=-40\end{cases}}\)
HM
1
29 tháng 6 2022
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow x^2-10x-2000=0\)
\(\Leftrightarrow x^2-10x+25-2025=0\)
\(\Leftrightarrow\left(x-5\right)^2=2025\)
=>x-5=45 hoặc x-5=-45
=>x=50 hoặc x=-40
21 tháng 2 2020
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\Rightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Giải rõ hơn nha
21 tháng 2 2020
\(\Leftrightarrow\frac{x^2-10x-29}{1971}+1+\frac{x^2-10x-27}{1973}+1-\frac{x^2-10x-1971}{29}-1-\frac{x^2-10x-1973}{27}-1=0\)
sai dấu r
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)=> \(\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)=>\(\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\) => \(\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)\)
=> \(x^2-10x-2000=0\)
Tự giải ra nhé hi hi
mik lm đc đến đoạn này rùi k bt lm