\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)
mình cần gấp mong các bạn giúp đỡ
Những câu hỏi liên quan
\(\left(3x-5\right).\left(-2x-7\right)=0\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(x^2-9=\left(1-4x\right)\left(x+3\right)\)
\(x^3-2x=-x^2+2\)
\(9x^2-16-x\left(3x+16\right)=0\)
\(\frac{2+4+...+2016+2018}{1019090}=-3x^2-4x\)
mình cần gấp mong các bạn giúp đỡ
\(\left(3x-5\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)
\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)
\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)
\(\Leftrightarrow3x^2+2+7x=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)
Không cần các bạn giải quyết hết, nhé! GẤP, GẤP, GẤPGiải các phương trình sau:a) x4 + 2x3 - 39x2 - 4x + 4 0b) (x + 4)4 + (x + 2)2 34c) frac{2x}{3x^2-5x+2}+frac{13x}{3x^2+x+2}6d) frac{x^4+3x^2+1}{x^3+x^2-x}3e) frac{1}{x^2}+frac{1}{left(x+1right)^2}15f) left(frac{x+1}{x-2}right)^2+frac{x-1}{x+3}12left(frac{x-2}{x+3}right)^2g) frac{2left(x+1right)}{3x^2+x}+frac{13left(x+1right)}{3x^2+7x+6}6
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Không cần các bạn giải quyết hết, nhé! GẤP, GẤP, GẤP
Giải các phương trình sau:
a) x4 + 2x3 - 39x2 - 4x + 4 = 0
b) (x + 4)4 + (x + 2)2 = 34
c) \(\frac{2x}{3x^2-5x+2}+\frac{13x}{3x^2+x+2}=6\)
d) \(\frac{x^4+3x^2+1}{x^3+x^2-x}=3\)
e) \(\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}=15\)
f) \(\left(\frac{x+1}{x-2}\right)^2+\frac{x-1}{x+3}=12\left(\frac{x-2}{x+3}\right)^2\)
g) \(\frac{2\left(x+1\right)}{3x^2+x}+\frac{13\left(x+1\right)}{3x^2+7x+6}=6\)
c,chia cả tử và mẫu cho x,sau đó đặt 3x+2/x=t
các câu còn lại hiện chưa giải đc vì chưa có giấy nháp,lúc nào rảnh mình chỉ cho cách làm
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Mọi người giải giúp mình bài này với:B2: Giải các PT sau:l) left(x-1right)left(5x+3right)left(3x-8right)left(x-1right)m) 2xleft(x-1right)x^2-1n) left(2-3xright)left(x+11right)left(3x-2right)left(2-5xright)o) frac{3}{7}x-1frac{1}{7}xleft(3x-7right)p) left(x-frac{3}{4}right)^2+left(x-frac{3}{4}right)left(x-frac{1}{2}right)0q) frac{1}{x}x+2left(frac{1}{x}+2right)left(x^2+1right)r) left(2x+3right)left(frac{3x+8}{2-7x}+1right)left(x-5right)left(frac{3x+8}{2-7x}+1right)s) left(x+2right)left(x-3right)l...
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Mọi người giải giúp mình bài này với:
B2: Giải các PT sau:
l) \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
m) \(2x\left(x-1\right)=x^2-1\)
n) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
o) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)
p) \(\left(x-\frac{3}{4}\right)^2+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)=0\)
q) \(\frac{1}{x}x+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
r) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
s) \(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
Các Pro giúp mình với !
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
Xem thêm câu trả lời
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
mình cần gấp mong các bạn giúp đỡ
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)
\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)
=> PT đúng với mọi x khác \(\pm5\)
Refund QB nhìn logic :V
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)
\(20x=2x^2+50\)
\(20x-2x^2-50=0\)
\(2\left(10x-x^2-25\right)=0\)
\(-x^2+10x+25=0\)
\(x^2-10x+25=0\)
\(x^2-2\left(x\right)\left(5\right)+5^2=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\Leftrightarrow x=5\)
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Xem thêm câu trả lời
Tìm x :
a) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
b) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
c) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
d) \(\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
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2) Giải phương trìnha) frac{x+1}{x-2}+frac{x-1}{x+2}frac{2left(x^2+2right)}{x^2-4}b) left(2x+3right).left(frac{3x+8}{2-7x}+1right)left(x-5right).left(frac{3x+8}{2-7x}+1right)3) Rút gọna) frac{2x-1}{x^3+1}+frac{2x}{x^2-x+1}+frac{-x}{x+1}+2b) frac{x+1}{2x-2}+frac{x^2+3}{2-2x^2}+frac{1}{1-x}-1,5c) left(frac{x^2}{x^3-4x}-frac{6}{3x-6}+frac{1}{x+2}right).frac{x+2}{6}d) left(frac{x}{xy-y^2}+frac{2x-y}{xy-x^2}right):frac{x^2-2xy+y^2}{x^2y-xy^2}e) [frac{1}{left(2x-yright)^2}+frac{2}{4x^2-y^2}-frac{1}{le...
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2) Giải phương trình
a) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
b) \(\left(2x+3\right).\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right).\left(\frac{3x+8}{2-7x}+1\right)\)
3) Rút gọn
a) \(\frac{2x-1}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{-x}{x+1}+2\)
b) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}+\frac{1}{1-x}-1,5\)
c) \(\left(\frac{x^2}{x^3-4x}-\frac{6}{3x-6}+\frac{1}{x+2}\right).\frac{x+2}{6}\)
d) \(\left(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}\right):\frac{x^2-2xy+y^2}{x^2y-xy^2}\)
e) \([\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}-\frac{1}{\left(2x+y\right)^2}].\frac{x^2+4xy+y^2}{16x}\)
Mn giúp mik vs mik đang cần gấp
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
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a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
b)\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
c)\(\frac{x +1}{x-2}+\frac{x-1}{x +2}=\frac{2\left(x^{2^{ }}+2\right)}{x^2-4}\)
d)(2x+3)\(\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
Mình mong các bạn giúp mình, mình xin cảm ơn những bạn đã giúp mình
\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
\(\frac{x}{2}-\left(\frac{3}{5}x-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
\(\Rightarrow\frac{5x-6x+26+14+7x}{10}=0\Rightarrow6x+40=0\Rightarrow x=-\frac{20}{3}\)
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\(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\left|x^2-3x\right|+\left|\left(x+1\right)\left(x-3\right)\right|=0\)
Mình cần gấp lắm ạ , Giúp mk làm bài này với ,Cảm ơn các bạn rất nhiều <3
Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
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