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Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0
a, ĐKXĐ: x\(\ne\) 1;-1;2
b, A= \(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{2x^2-2x}{2\left(x+1\right)\left(x-1\right)}+\frac{2x+2}{2\left(x+1\right)\left(x-1\right)}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-2}{x+1}\)
=\(\frac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{x-2}{x-1}\)
c, Khi x= -1
→A= \(\frac{-1-2}{-1-1}\)
= -3
Vậy khi x= -1 thì A= -3
Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^
a,ĐKXĐ:x#1; x#-1; x#2
b,Ta có:
A=\(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{x\left(x-1\right)2}{\left(x+1\right)\left(x-1\right)2}+\frac{\left(x+1\right)2}{\left(x-1\right)\left(x+1\right)2}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{x-2}\)
=\(\frac{2x^2-2x+2x+2+4x}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{x-2}{x+1}\)
c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả
d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên
\(\Leftrightarrow x-2⋮x+1\)
\(\Leftrightarrow x+1-3⋮x+1\)
Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)
Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
Bài 1:
a) x≠2x≠2
Bài 2:
a) x≠0;x≠5x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.
=> x=−5x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). dấu = xảy ra khi x2=0
=> x=0
Vậy \(B_{min}\Leftrightarrow x=0\)
ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)
dấu = xảy ra khi \(x+1=0\)
\(\Rightarrow x=-1\)
Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
Bài 2 :
a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)
b) \(A=\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\cdot\frac{x^2-4x+4}{4}\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)
\(A=\frac{x-2}{x+2}\)
c) Thay x = 4 ta có :
\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)
Vậy.........
\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)
\(\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)
\(=\left(5x\right)^3-2^3\)
\(=125x^3-8\)
A=x3/x2--4.x+2/x-x-4xx-4/xx-2
Điều kiện x \(\ne\)+-2
Ý b c tự làm
a) x2 - 4 ≠ 0 ⇔ (x + 2)(x - 2) ≠ 0
ĐKXĐ: x ≠ - 2 và x ≠ 2