Tìm x :
a) ( 2x + 3 ) 2 = \(\frac{9}{21}\)
b) ( 3x - 1 ) 3 = \(\frac{-8}{27}\)
c) \(|2x+1|\)- 3 = 6
giải phương trình
a) \(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-\sqrt{2x-1}}=4\)
b) \(4x^2+3x+3=4x\sqrt{x+3}+2\sqrt{2x-1}\)
c) \(\sqrt{x-4}+\sqrt{6-x}=x^2-11x+27\)
d) \(\sqrt{13x^2-6x+10}+\sqrt{5x^2-13x+\frac{17}{2}}+\sqrt{17x^2-48x+36}=\frac{1}{2}\left(36x-8x^2-21\right)\)
e) \(\sqrt{\frac{6}{3-x}}+\sqrt{\frac{8}{2-x}}=6\)
1) giải phương trình:
a) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x+5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{7x+10}{x+1}\left(x^2-x-2\right)-\frac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\frac{2x+5}{x+3}+1=\frac{4}{x^2+2x-3}-\frac{3x-1}{1-x}\)
d) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}+\frac{6}{9-x^2}=0\)
e) \(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
f) \(\frac{1+\frac{x}{x+3}}{1-\frac{x}{x+3}}=3\)
Thực hiện phép tính:
a)\(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)
b)\(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)
c)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
d)\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
e)\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
f)\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
g)\(\frac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\frac{2}{x^2+3}+\frac{1}{x+1}\)
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
Bài 8: Tìm số nguyên x biết
a) \(\left(\frac{-12}{27}+\frac{2}{3}\right)+\frac{-2}{9}\le x\le\left(\frac{11}{7}+\frac{2}{5}\right)+\frac{7}{5}+\frac{3}{7}\) b\(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
c)\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
tìm x biết: a)\(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
b)
\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
c)\(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{2}{9-6x}-\frac{3}{2}\)
a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
Tìm x thỏa mãn:
a) \(2\left|\frac{2}{3}-x\right|+\frac{1}{4}=\frac{3}{4}\)
b) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
c) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
rút gọn
a, (2x-1) (3x+5)-2(-4x+1)2
b, \(\frac{x^2-16}{4x-x^2}\)
c, \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
d, (x-1)3-(x+1)3+6(x+1) (x-1)
e, (2x+7)2-(4x+14) (2x-8)+(8-2x)2
a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7
b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)
c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)
d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)
= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)
= -2(x2 - 2x + 1 + x2 - 1 + x2 + 2x + 1) + 6x2 - 6
= -2(3x2 + 1) + 6x2 - 6
= -6x2 - 2 + 6x2 - 6
= -8
e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2
= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2
= (2x + 7 - 2x + 8)2
= 152 = 225
Tìm x biết
a) x+2x+3x+4x+...+100x=-213
b)\(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
c)3(x-2)+2(x-1)=10
d)\(\frac{x+1}{3}=\frac{x-2}{4}\)
e)\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
f)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a) x + 2x + 3x + ... +100x = -213
=> x . (1 + 2 + 3 +... + 100) = - 213
=> x . 5050 = -213
=> x = - 213 : 5050
=> x = -213/5050
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)
=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)
=> \(x.\frac{1}{4}=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{1}{4}\)
=> \(x=\frac{2}{3}\)
c) 3(x-2) + 2(x-1) = 10
=> 3x - 6 + 2x - 2 = 10
=> 3x + 2x - 6 - 2 = 10
=> 5x - 8 = 10
=> 5x = 10 + 8
=> 5x = 18
=> x = 18:5
=> x = 3,6
d) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> \(4\left(x+1\right)=3\left(x-2\right)\)
=>\(4x+4=3x-6\)
=> \(4x-3x=-4-6\)
=> \(x=-10\)