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Minh Anh Doan
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Tiên Nguyễn
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Nguyễn Lê Phước Thịnh
27 tháng 10 2021 lúc 19:55

a: \(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(=\dfrac{2\sqrt{x}+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}}{x-1}\)

ILoveMath
27 tháng 10 2021 lúc 19:56

\(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{x-1}\)

\(\Rightarrow P=\dfrac{\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}-1}\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{2+2\sqrt{2}}\)

\(\Rightarrow P=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}\)

\(\Rightarrow P=\dfrac{1}{2}\)

 

Minh Anh Doan
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Lấp La Lấp Lánh
27 tháng 9 2021 lúc 13:43

Bài 1:

a) \(=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1=\left(x+1\right)\left(x+3\right)\)

f) \(=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)

h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

k) \(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3-x-1\right)\)

m) \(=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

Bài 2:

a) \(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

e) \(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

f) giống câu a

g) \(=x^2-2xy=x\left(x-2y\right)\)

i) \(=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

k) \(=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

l) \(=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\)

Minh Anh Doan
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ILoveMath
15 tháng 11 2021 lúc 15:31

a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)

b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2}{2x+1}\)

c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)

 

Minh Anh Doan
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Monkey D. Luffy
14 tháng 11 2021 lúc 15:47

Bài 6

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)

Nguyễn Hoàng Minh
14 tháng 11 2021 lúc 15:50

Bài 7:

\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)

Minh Hiếu
14 tháng 11 2021 lúc 15:49

Điều kiện (x≠5, x≠-5)

\(P=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

    \(=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}+\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

    \(=\dfrac{x-5+2\left(x+5\right)-2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)

    \(=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

Minh Anh Doan
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Minh Anh Doan
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Minh Anh Doan
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Monkey D. Luffy
15 tháng 11 2021 lúc 10:12

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Như Tâm
15 tháng 11 2021 lúc 10:15

Bào quan riboxom trong chất tế bào có chức năng gì? 

ILoveMath
15 tháng 11 2021 lúc 10:16

a) \(=\left(x^2+x\right)+\left(4x+4\right)=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

b) \(=\left(x^2+2x\right)-\left(3x+6\right)=x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)

c) \(=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

d) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left[\left(x^2+5x\right)-\left(2x+10\right)\right]=3\left[x\left(x+5\right)-2\left(x+5\right)\right]=3\left(x-2\right)\left(x+5\right)\)

e) \(=-\left(3x^2-5x-2\right)=-\left[\left(3x^2-6x\right)+\left(x-2\right)\right]=-\left[3x\left(x-2\right)+\left(x-2\right)\right]=-\left(3x+1\right)\left(x-2\right)\)

f) \(x^2-7x+6=\left(x^2-x\right)-\left(6x-6\right)=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

h) \(=4\left(x^2-9x+14\right)=4\left[\left(x^2-7x\right)-\left(2x-14\right)\right]=4\left[x\left(x-7\right)-2\left(x-7\right)\right]=4\left(x-2\right)\left(x-7\right)\)

i) \(=3\left(3x^2-8x+5\right)=3\left[\left(3x^2-3x\right)-\left(5x-5\right)\right]=3\left[3x\left(x-1\right)-5\left(x-1\right)\right]=3\left(x-1\right)\left(3x-5\right)\)

k) \(=-\left(2x^2+5x+2\right)=-\left[\left(2x^2+4x\right)+\left(x+2\right)\right]=-\left[2x\left(x+2\right)+\left(x+2\right)\right]=-\left(x+2\right)\left(2x+1\right)\)

l) \(=\left(x^2-5xy\right)-\left(2xy-10y^2\right)=x\left(x-5y\right)-2y\left(x-5y\right)=\left(x-5y\right)\left(x-2y\right)\)

m) \(=\left(x^2-2xy\right)-\left(xy-2y^2\right)=x\left(x-2y\right)-y\left(x-2y\right)=\left(x-2y\right)\left(x-y\right)\)

n) \(=\left(x^2-3xy\right)+\left(xy-3y^2\right)=x\left(x-3y\right)+y\left(x-3y\right)=\left(x+y\right)\left(x-3y\right)\)

Minh Anh Doan
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Edogawa Conan
27 tháng 9 2021 lúc 19:28

Bài 1.

a, PTPƯ: kẽm + axit clohidric → kẽm clorua + hidro

b, Theo ĐLBTKL ta có:

 \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

c, Ta có: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=27,2+0,4-13=14,6\left(g\right)\)

Bài 2:

a, PTPƯ: metan + oxi → cacbon dioxit + hơi nước

b, Theo ĐLBTKL ta có: 

 \(m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\)

c, Ta có: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=132+108-48=192\left(g\right)\)  

𝓓𝓾𝔂 𝓐𝓷𝓱
27 tháng 9 2021 lúc 19:26

 PT chữ: Metan + Oxi \(\xrightarrow[]{t^o}\) Cacbon đioxit + Nước

Bảo toàn khối lượng: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=192\left(g\right)\)

𝓓𝓾𝔂 𝓐𝓷𝓱
27 tháng 9 2021 lúc 19:28

Bài 1:

PT chữ: Kẽm + Axit clohidric \(--->\) Kẽm clorua + Hidro

Bảo toàn khối lượng: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=14,6\left(g\right)\)