Rút gọn:
P=(a+1).(b-2)-(ab-2)
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T
9 tháng 10 2023
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\left(a>0;a\ne1\right)\)
\(=\left(\dfrac{\sqrt{a}\cdot\sqrt{a}}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left[\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\left[\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\dfrac{\left(a-1\right)^2}{\left(2\sqrt{a}\right)^2}\cdot\dfrac{-4\sqrt{a}}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{-4\sqrt{a}}{a-1}\)
\(=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
\(=\dfrac{1-a}{\sqrt{a}}\)
ND
Rút gọn:
P=\(\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)
1
7 tháng 12 2021
\(P=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
AT
19 tháng 7 2021
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{1-x}\right)\left(\dfrac{2}{\sqrt{x}}-\dfrac{2}{x}\right)\left(x>0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{2\sqrt{x}-2}{x}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2+2\sqrt{x}\left(\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{2\left(\sqrt{x}-1\right)}{x}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{2\left(\sqrt{x}-1\right)}{x}=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{2\left(\sqrt{x}-1\right)}{x}\)
\(=\dfrac{6}{\sqrt{x}}\)
19 tháng 7 2021
Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{1-x}\right)\cdot\left(\dfrac{2}{\sqrt{x}}-\dfrac{2}{x}\right)\)
\(=\dfrac{x-2\sqrt{x}+1+2x+2\sqrt{x}+3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{2\sqrt{x}-2}{x}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{2\sqrt{x}-2}{x}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)\cdot2\cdot\left(\sqrt{x}-1\right)}{x\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}}\)
LA
8 tháng 3 2018
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)
\(\Leftrightarrow bc=-ac-ca\Leftrightarrow a^2+2bc=a^2+bc-ca-ab\)
\(\Leftrightarrow a^2+2bc=\left(a-c\right)\left(a-b\right)\)
Tương tự với 2 phân số còn lại rồi quy đồng
20 tháng 1 2019
(a+b+c)^2=a^2+b^2+c^2
<=>a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2
<=.2ab+2ac+2bc=0
<=>ab+ac+bc=0
<=>bc=-ab-ac
Ta có : a^2/(a^2+2bc)=a^2/(a^2+bc+bc)=a^2/(a^2+bc-ab-ac)=a^2/[a(a-b)-c(a-b)]=a^2/(a-b)(a-c) (1)
chứng minh tương tự ta được: b^2/(b^2+2ac)=b^2/(b-a)(b-c) (2)
c^2/(c^2+2ab)=c^2/(c-a)(c-b) (3)
Cộng vế với vế của (1)(2)(3) ta được :
a^2/(a^2+2bc)+b^2/(b^2+2ac)+c^2/(c^2+2ab)=a^2/(a-b)(a-c)+b^2/(b-a)(b-c)+c^2/(c-a)(c-b)
hay P=a^2/(a-b)(a-c)-b^2(b-c)(a-b)+c^2/(a-c)(a-b)
=a^2(b-c)/(a-b)(a-c)(b-c)-b^2(a-c)/(a-b)(a-c)(b-c)+c^2(a-b)/(a-b)(a-c)(b-c)
=(a^2b-a^2c-b^2a+b^2c+c^2a-c^2b)/(a-b)(a-c)(b-c)
=(a^2b+b^2c-a^2c-c^2b-b^2a+c^2a)/(a-b)(a-c)(b-c)
=[b(a^2+bc)-c(a^2+bc)-a(b^2-c^2)]/(a-b)(a-c)(b-c)
=[(b-c)(a^2+bc)-a(b-c)(b+c)]/(a-b)(a-c)(b-c)
=[(b-c)(a^2+bc-ab-ac)]/(a-b)(a-c)(b-c)
={(b-c)[a(a-b)-c(a-b)]}/(a-b)(a-c)(b-c)
=(b-c)(a-c)(a-b)/(a-b)(a-c)(b-c)
=1
Vậy P=1
AT
19 tháng 7 2021
\(P=\left(1+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}+1}{x\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{x+2\sqrt{x}+1}{x+\sqrt{x}+1}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{x+\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=x-1\)
19 tháng 7 2021
Ta có: \(P=\left(1+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}+1}{x\sqrt{x}-1}\)
\(=\dfrac{x+2\sqrt{x}+1}{x+\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=x-1\)
26 tháng 11 2021
\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
\(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)
\(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)
\(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)
Đề lỗi rồi chứ mình ko rút gọn đc nữa