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\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
a)Đkxđ x≠\(\frac{5}{4}\)
Ta có để \(\frac{2x+3}{4x-5}\)=0=>2x+3=0=>x=\(\frac{3}{2}\)(thỏa mãn)
b)Ta có \(x^2-4x+3=x^2-3x-x+3\)
=x(x-3)-(x-3)
=(x-1)(x-3)
=>Đkxđ x≠1,3
để bài b)=0 duy ra (x-1)(x-2)=0
=>x=1,x=2 đối chiếu đkxđ có x=2 (t/mãn)
c)phân thức tương đương:\(\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
= \(\frac{x+1}{x-1}\)
=>Đkxđ x≠1
Để x+1/x-1=0=>x+1=0
=>x=-1(t/mãn)
d) phân thức tương đương
\(\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+5\right)}\)
=\(\frac{x+2}{x+5}\)=>x≠-5
để phân thức đạt 0 suy ra x+2=0
=>x=-2
e)phân thức tương đương
\(\frac{x\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)\left(x+1\right)}\)
=\(\frac{x+4}{x+1}\)
Đkxđ x khác -1
Để phân thức đạt GT là 0 x+4=0=>x=-4
g)\(\frac{\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x^2+x+3\right)}\)
=\(\frac{\left(x+1\right)^2}{x^2+x+3}\)
vì\(x^2+x+3>0\)(Dễ dàng chứng minh)
=>xϵR
Để phân thức đạt gt là 0 => \(\left(x+1\right)^2=0=>x=-1\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
Làm ngắn gọn thôi nhé :v
\(A=\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)
\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)
\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x+2}{x-3}\)
\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)
\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)
\(B=\frac{\left(-x-2\right)\left(x+2\right)\left(x+2\right)\left(x-2\right)}{\left(-x-2\right)\left(x-2\right)\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{x+2}{x-2}\)
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{10x}{-x^2+9}\)
\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)
\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)
\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)
\(D=\frac{51x-15}{2x^3-18x}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)
\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)
\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(E=\frac{10x^2+10}{x^4-2x+1}\)
a) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\)
\(=\frac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)
c) \(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}\)
\(=\frac{2x\left(x-2\right)^2}{x\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x\left(x-2\right)}{x\left(x+2\right)}\)
\(=\frac{2x^2-4x}{x^2+2x}\)
d) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)
\(=\frac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x}{x+y}\)
a) Để \(\frac{2x+3}{4x-5}=0\)
=> 2x + 3 = 0
x = -3/2
b) Để \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-3\right).\left(x-1\right)}=\frac{x+2}{x-3}=0\)
=> x + 2 = 0=> x = -2
c) để \(\frac{x^2-1}{x^2-2x+1}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-1\right)^2}=\frac{x+2}{x-1}=0\)
=> x + 2 = 0 => x = -2
d) để \(\frac{x^2-4}{x^2+3x-10}=\frac{\left(x+2\right).\left(x-2\right)}{\left(x-2\right).\left(x+5\right)}=\frac{x+2}{x+5}=0\)
=> ...
e) để \(\frac{x^3-16x}{x^3-3x^2-4x}=\frac{x.\left(x-4\right).\left(x+4\right)}{x.\left(x-4\right).\left(x+1\right)}=\frac{x+4}{x+1}=0\)
=> ....