I : giải các phương trình
a) \(\dfrac{x-1}{a}+\dfrac{2}{x+1}=\dfrac{x^2-2ax-1}{a\left(x+1\right)}\)(ẩn x )
b) \(4x^2+4mx+m^2-25=0\)
help me !!!
I : Giải các phường trình sau
a) \(\left(3x-2\right)\left(\dfrac{2\left(x+3\right)}{7}-\dfrac{4x-3}{5}\right)=0\)
b) \(\left(x-\dfrac{3}{4}\right)^2+\left(x-\dfrac{3}{4}\right)\left(x-\dfrac{1}{2}\right)=0\)
c) \(\dfrac{12}{9-x^2}+\dfrac{2}{x-3}+\dfrac{3}{x+3}=1\)
d) \(\dfrac{1}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
help me
1, giải phương trình
a, 2( x - 0,5 ) + 3 = 0,25(4x - 1 )
b , 2.\(\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0,5\right)+2\)
2, giải các phương trình
a, \(\dfrac{3x-2}{2}=\dfrac{1-2x}{3}\)
b, \(\dfrac{x-1}{3}+2=3-\dfrac{2x+5}{4}\)
c, \(\dfrac{x-1}{5}+x=\dfrac{x+1}{7}\)
d, 2(x - 2,5 ) = 0,25 +\(\dfrac{4x-3}{8}\)
các bạn ơi ! giúp mik với ! mai kiểm tra rồi
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
a. 2( x - 0,5 ) + 3 = 0,25(4x - 1 )
(=) 2x - 1 + 3 = x - 0,25
(=) 2x - x = 1 - 3 - 0,25
(=) x = -2,25
b. 2. (x−14) −4 = −6 (−13x+0,5) +2
(=) 2x - 0.5 - 4 = 2x - 3 + 2
(=) 2x - 2x = 0,5 + 4 -3 +2 ( vô lí )
Vậy phương trình này vô nghiệm
Giải các phương trình sau :
A) \(\dfrac{7x-1}{2}\)+2x=\(\dfrac{16-x}{3}\)
B)\(\dfrac{x+1}{x-2}\)+\(\dfrac{x-1}{x+2}\)=\(\dfrac{2\left(x^2+2\right)}{x^2-4}\)
a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)
\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)
khử mẫu
=> (7x-1).3+12x=(16-x).2
=>21x-3+12x=-2x+32
=>21x-3+12x+2x-32=0
=>35x-35=0
b)\(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
ĐKXĐ: x khác +-2
\(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
khử mẫu
(x+1).(x+2)+(x-1)(x-2)=2x2+4
=>x2+x+2+x+2+x2-2x-x+2=2x2+4
=>x2+x+2+x+2+x2-2x-x+2-2x2-4=0
=>(x2+x2-2x2)+(x+x-2x-x)+(2+2+2-4)=0
=>-x+2=0
=>-x=-2
=>x=2(loại)
vậy pt vô nghiệm
B1: Giải các phương trình sau
1) \(\left(3x-2\right)\left(\dfrac{2\left(x+3\right)}{7}-\dfrac{4x-3}{5}\right)=0\)
2) \(\left(x-\dfrac{3}{4}\right)^2+\left(x-\dfrac{3}{4}\right)\left(x-\dfrac{1}{2}\right)=0\)
3) \(\dfrac{12}{9-x^2}+\dfrac{2}{x-3}+\dfrac{3}{x+3}=1\)
4) \(\dfrac{1}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
các bạn ơi giúp mình với
mik cảm ơn !!!
Giải các phương trình và bất phương trình sau
a)\(\left|x-9\right|\) \(=2x+5\)
b) \(\dfrac{1-2x}{4}\) \(-2\) ≤ \(\dfrac{1-5x}{8}\) + x
c)\(\dfrac{2}{x-3}\)\(+\dfrac{3}{x+3}\)\(=\dfrac{3x+5}{x^2-9}\)
|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
Ta có:
\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)
\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)
\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)
I : Giải các phương trình
a) \(\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\)
b) \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
c) \(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\)
help me
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
1. Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\left(a+b\right)^2=10+2ab\\\left(a+b\right)\left(a-\dfrac{2}{ab}\right)=\dfrac{4}{3}\end{matrix}\right.\)
2.Giải phương trình:
\(\sqrt{\sqrt{2}-1-x}+\sqrt[4]{x}=\dfrac{1}{\sqrt[4]{2}}\)
B1: Tính:
\(B=\dfrac{4.\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
B2: Xác định a, b, c:
a, \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{1-2}+\dfrac{c}{n+2}\) với mọi x khác 0, x khác \(\pm2\)
b, \(\dfrac{1}{x^3-1}=\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)
Help me!!!
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
Bài 2.
Sửa đề
a) \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)
Giải
Ta sẽ phân tích vế phải
VP = \(\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)
VP = \(\dfrac{a\left(x^2-4\right)+bx\left(x+2\right)+cx\left(x-2\right)}{x\left(x^2-4\right)}\)
VP = \(\dfrac{ax^2-4a+bx^2+2bx+cx^2-2cx}{x\left(x^2-4\right)}\)
VP = \(\dfrac{x^2\left(a+b+c\right)+2x\left(b-c\right)-4a}{x\left(x^2-4\right)}\)
Tương tự , ta cũng sẽ phân tích VT
VT = \(\dfrac{2x.5-4}{x\left(x^2-4\right)}\)
Đồng nhất hai VT và VP , ta có :
\(x^2\left(a+b+c\right)+2x\left(b-c\right)-4a=2.5x-4\)
* a + b + c = 0 => 1 + c + 5 + c = 0 => 2c = - 6 => c = - 3
* b - c = 5 => b = c + 5 => b = - 3 + 5 => b = 2
* a = 1
Vậy , a = 1 ; b = 2 ; c = -3
b) Ta sẽ phân tích VP
VP = \(\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)
VP = \(\dfrac{a\left(x^2+x+1\right)+\left(bx+c\right)\left(x-1\right)}{x^3-1}\)
VP = \(\dfrac{ax^2+ax+a+bx^2-bx+cx-c}{x^3-1}\)
VP = \(\dfrac{x^2\left(a+b\right)+x\left(a-b+c\right)+a-c}{x^3-1}\)
Đồng nhất VP và VT , ta được :
\(x^2\left(a+b\right)+x\left(a-b+c\right)+a-c=1\)
* a + b = 0 => a = - b => b = \(-\dfrac{1}{3}\)
* a - b + c = 0 => a + a + a - 1 = 0 => 3a = 1 => a = \(\dfrac{1}{3}\)
* a - c = 1 => c = a - 1 => c = \(\dfrac{1}{3}\) - 1 = \(-\dfrac{2}{3}\)
Vậy , a = \(\dfrac{1}{3}\) ; b = \(-\dfrac{1}{3}\); c = \(-\dfrac{2}{3}\)
Bài 1 bạn Giang làm rồi thì thôi nhé
Kiểm tra giùm mk câu a bài 2 nha!!! ĐỀ BÀI!!!
giải các phương trình sau
a, 4x- 2(1-x)= 5(x-4)
b, \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
c, \(\left(x+2\right)^2-3\left(x+2\right)=0\)
d,\(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
d: Ta có: \(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)
\(\Leftrightarrow x^2-25+x^2-3x=x-25\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)