Giải phương trình
a, \(x^4+2x^3-4x^2-5x-6=0\)
b, \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+\dfrac{1}{100.110}\)
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28 tháng 1 2018
b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)
<=> \(\dfrac{7}{2}-\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{9}{2}\)
<=> \(\dfrac{15}{4}-\dfrac{x}{5}-\dfrac{9}{2}=0\)
<=> \(\dfrac{x}{5}=\dfrac{5}{4}\)
<=> x = 6,25
Vậy,...
c) ( x + 2)( x + 3)( x - 5)( x - 6) = 180
<=> ( x + 2)( x - 5)( x + 3)( x - 6) = 180
<=> ( x2 - 3x - 10 )( x2 - 3x - 18 ) = 180
Đặt : x2 - 3x - 14 = a , ta có :
( a + 4)( a - 4) = 180
<=> a2 - 16 - 180 = 0
<=> a2 - 196 = 0
<=> ( a - 14)( a + 14 ) = 0
<=> a = 14 hoặc a = -14
* Với , a = 14 , ta có :
x2 - 3x - 14 = 14
<=> x2 - 3x - 28 = 0
<=> x2 - 7x + 4x - 28 = 0
<=> x( x - 7) + 4( x - 7) = 0
<=> ( x + 4)( x - 7) = 0
<=> x = -4 hoặc : x = 7
* Với : a = -14 , ta có :
x2 - 3x - 14 = -14
<=> x( x - 3) = 0
<=> x = 0 hoặc : x = 3
Vậy,...
30 tháng 6 2022
b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
=>x=3/11+20/55=3/11+4/11=7/11
c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
\(\Leftrightarrow x-100=1\)
hay x=101
Q
27 tháng 4 2017
Tìm x, biết: \(\left ( \frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110} \right ).x = \frac{1}{1.11}+\frac{1}{1.12}+...+\frac{1}{100.110}\)- Trường Toán Trực tuyến Pitago – Giải pháp giúp em học toán vững vàng!
20 tháng 4 2020
Câu hỏi của Huỳnh Ngọc Cẩm Tú - Toán lớp 6 - Học toán với OnlineMath
NA
1
5 tháng 3 2019
Ta có:
$(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$
$\Leftrightarrow \frac{1}{100}\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=\frac{1}{10}\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
$\Leftrightarrow \left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
Đặt $A=\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )+\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}$
Thay vào phương trình, ta có:
$\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )$
$\Leftrightarrow x=10$
28 tháng 8 2021
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
15 tháng 5 2023
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
NH
1
5 tháng 3 2020
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+\frac{100}{3\cdot103}+...+\frac{100}{100\cdot110}\right)x=10\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+....+\frac{1}{10}-\frac{1}{110}\right)x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{110}\right)\right]x=10\)\(\left[\left(1+\frac{1}{2}+....+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{110}\right)\right]\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\right]x=10\)
\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow x=10\)
a, ⇔ x4 - 2x3 + 4x3 - 8x2 + 4x2 - 8x + 3x - 6 = 0
⇔ (x - 2)(x3 + 4x2 + 4x + 3) = 0
⇔ (x - 2)(x3 + 3x2 + x2 + 3x + x + 3) = 0
⇔ (x - 2)(x + 3)(x2 + x + 1) = 0 mà x2 + x + 1 > 0 ∀ x
⇔ \(\left\{{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình S = {2; -3}