x²-x+1
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LARGE 7)Gfrac{sqrt{x}+1}{x-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2sqrt{x}+2}{x-1}Gfrac{sqrt{x}+1}{sqrt{x}-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2(sqrt{x}+1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{sqrt{x}+1}{x-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2}{sqrt{x}-1}Gfrac{(sqrt{x}+1)^2+(sqrt{x}-1)^2-2(sqrt{x}+1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2x-2sqrt{x}}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2sqrt{x}(sqrt{x}-1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2sqrt{x}}{sqrt{x}+1}8)H(frac{1}{sqrt{x}-1}+frac{sqrt{x}}{x-1}):(frac{sqrt{x}}{sqrt{x}-1}-...
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\[\LARGE 7)G=\frac{\sqrt{x}+1}{x-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{x-1}\\G=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{\sqrt{x}+1}{x-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}\\G=\frac{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2-2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2x-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2\sqrt{x}}{\sqrt{x}+1}\\8)H=(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}):(\frac{\sqrt{x}}{\sqrt{x}-1}-1)\\H=(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}):(\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}-1})\\H=\frac{\sqrt{x}+1+\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\\H=\frac{2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{1}{\sqrt{x}-1}\\H=\frac{2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}.(\sqrt{x}-1)\\H=\frac{2\sqrt{x}}{\sqrt{x}+1}\]
1. x+1/x-1 - x-1/x+1 = 4/x2-1
2. x+2/x-2 + x/x+2 = 2
3. 2x/x-1 - 1/x+2 = 2
4. x/x2-25 - 1-x/x-5 = 1/x+5
5. x-1/x+2 - 9/x2-4 = -3/x-2
6. 3x-3/x2-9 - 1/x-3 = x+1/x+3
Xem chi tiết
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
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1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 =
1 nhé bạn
1
(số nào nhân với 1 bằng chính nó)
= 1
#Học tốt!!!
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1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 2 x3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13 x 14 x 15 =
the only
jjjjjji8hgjkoljjimkkpllkkkkkkllkkkkklkI-ol0loookkkkokpplplkoo
hack nè
1,307,674,368,000
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1/x-1-x^3-x/x^2+1(x/x^2-2x+1-1/x^2-1)
[2/(x+1)^3.(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
(x/x^2-36-x-6/x^2+6x):2x-6/x^2+6x+x/6-x
giúp mik với ;-; mik cần gấp
A1= \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)với x≠1, x≥ 0
A2= \(\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right]:\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)với x≥0, x≠1 và -1
\(A_1=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A_2=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x-\sqrt{x}+1}{x+1}\\ A_2=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-\sqrt{x}+1}\\ A_2=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
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1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 5 = ?
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tìm xa x² x 1 4x² 8x 4 0b x² x ² x² x 6c x 1 x² x 1 x 1 x² x 1 2 x 1 x 1 4 2 x 2 ²
Xem chi tiết
7 tháng 7 2021 lúc 10:49
Rút gọn
a) \((\dfrac{2x^2+3x}{x^3+1}+\dfrac{1}{x^2-x+1}).\dfrac{x^2-x+1}{x}\)
b) \(\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right):\left(\dfrac{x+1}{x-2}-\dfrac{x+2}{x-1}\right)\)
c) \(\left(\dfrac{1}{x}+\dfrac{x}{x+1}\right).\dfrac{x^2+x}{x}\)
Lời giải:
a. ĐKXĐ: $x\neq 0;-1$
\(=\left(\frac{2x^2+3x}{(x+1)(x^2-x+1)}+\frac{x+1}{(x+1)(x^2-x+1)}\right).\frac{x^2-x+1}{x}\)
\(=\frac{2x^2+3x+x+1}{(x+1)(x^2-x+1)}.\frac{x^2-x+1}{x}=\frac{2x^2+4x+1}{x(x+1)}\)
b. ĐKXĐ: $x\neq 0; 1;2$
\(=\frac{x-(x-1)}{x(x-1)}:\frac{(x+1)(x-1)-(x-2)(x+2)}{(x-2)(x-1)}=\frac{1}{x(x-1)}:\frac{3}{(x-2)(x-1)}\)
\(=\frac{1}{x(x-1)}.\frac{(x-2)(x-1)}{3}=\frac{x-2}{3x}\)
c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)
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