Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
1) Ta có : \(4x+20=0\)
=> \(x=-\frac{20}{4}=-5\)
Vậy phương trình có tập nghiệm là \(S=\left\{-5\right\}\)
2) Ta có : \(3x+15=30\)
=> \(3x=15\)
=> \(x=5\)
Vậy phương trình có tập nghiệm là \(S=\left\{5\right\}\)
3) Ta có : \(8x-7=2x+11\)
=> \(8x-2x=11+7=18\)
=> \(6x=18\)
=> \(x=3\)
Vậy phương trình có tập nghiệm là \(S=\left\{3\right\}\)
4) Ta có : \(2x+4\left(36-x\right)=100\)
=> \(2x+144-4x=100\)
=> \(-2x=-44\)
=> \(x=22\)
Vậy phương trình có tập nghiệm là \(S=\left\{22\right\}\)
5) Ta có : \(2x-\left(3-5x\right)=4\left(x+3\right)\)
=> \(2x-3+5=4x+12\)
=> \(-2x=10\)
=> \(x=-5\)
Vậy phương trình có tập nghiệm là \(S=\left\{-5\right\}\)
1) 4x+20=0
\(\Leftrightarrow\) 4x=-20
\(\Leftrightarrow\) x=-5
Vậy pt trên có tập nghiệm là S={-5}
2) 3x+15=30
\(\Leftrightarrow\) 3x=15
\(\Leftrightarrow\) x=5
Vậy pt trên có tập nghiệm là S={5}
3) 8x-7=2x+11
\(\Leftrightarrow\) 8x-2x=11+7
\(\Leftrightarrow\) 6x=18
\(\Leftrightarrow\) x=3
Vậy pt trên có tập nghiệm là S={3}
4) 2x+4(36-x)=100
\(\Leftrightarrow\) 2x+144-4x=100
\(\Leftrightarrow\) -2x+144=100
\(\Leftrightarrow\) -2x=-44
\(\Leftrightarrow\) x=22
Vậy pt trên có tập nghiệm là S={22}
5) 2x-(3-5x)=4(x+3)
\(\Leftrightarrow\) 2x-3+5x=4x+12
\(\Leftrightarrow\) 2x+5x-4x=12+3
\(\Leftrightarrow\) 3x=15
\(\Leftrightarrow\) x=5
Vậy pt trên có tập nghiệm là S={5}
6) 3x(x+2)=3(x-2)2
\(\Leftrightarrow\) 3x2+6x=3(x2-2x.2+22)
\(\Leftrightarrow\) 3x2+6x=3x2-12x+12
\(\Leftrightarrow\) 3x2-3x2+6x+12x=12
\(\Leftrightarrow\) 18x=12
\(\Leftrightarrow\) x=\(\frac{2}{3}\)
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
.....
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
.....
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
ban lam not ho minh hai cau cuoi nha