Câu a)

a+ 2a-ab=7

3a-ab=7

a(3-b)=7

Mà 7= (-1)x(-7) = 1x7= 7x1

Suy ra 

Vậy ...................... (Kết Luận)

Câu b)

2a + 3ab - ab = -2

2a + 2ab = -2

2a(1+b)=-2

Mà -2= (-1)x2 = (-2)x1= 2x(-1) = 1x(-2)

Suy ra có bản như sau

Vậy ...................

Mấy câu kia làm cũng giống vậy nha..............

@ngonhuminh

a: \(B-2=\dfrac{4\sqrt{a}-4a-1}{2a+1}=\dfrac{-\left(2\sqrt{a}-1\right)^2}{2a+1}< 0\)

=>B<2

b: Để 1/D là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+3⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\{1;-1;3\right\}\)

hay \(x\in\left\{4;0;16\right\}\)

\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)

\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)

\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

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

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)

a,\(a^2\left(a+1\right)+2a\left(a+1\right)=\left(a^2+2a\right)\left(a+1\right)\)

\(=a\left(a+2\right)\left(a+1\right)⋮3⋮2\)

                                               \(⋮6\left(ĐPCM\right)\)

b,\(a\left(2a-3\right)-2a\left(a+1\right)\)

\(=2a^2-3a-2a^2-2a\)

\(=-5a⋮5\left(ĐPCM\right)\)

1:

a: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2zx+2yz\)

b: \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy+2xz-2yz\)

c: \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy-2xz+2yz\)

Bài 2: tất cả đều ở dạng tích rồi mà

\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)

\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(a\ge0\)

\(\Rightarrow a\in\left\{0;2;3\right\}\)

Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)

\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)

\(=\dfrac{2}{a-1}\)

Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)

hay \(a\in\left\{2;0;3\right\}\)

a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)

Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)

nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)

Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)

mà a+b+c=2 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)

Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)

b) Ta có: 2a=3b=5c

nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)

mà a+b-c=3

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)

Do đó: 

\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)

Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)

Xét \(M=\left(2a+1\right)^2+\left(2a+3\right)^2=4a^2+4a+1+4a^2+12a+9=8a^2+16a+10.\)

\(M=8\left(a+1\right)^2+2=2\left(4\left(a+1\right)^2+1\right)\)

4(a + 1)² + 1 là 1 số lẻ => M chia hết cho 2 mà không chia hết cho 4.

Hay M khi phân tích ra thừa số nguyên tố thì thừa số 2 có số mũ lẻ (=1) nên M không phải là số chính phương.

=> \(\sqrt{M}\)là số vô tỷ, hay \(\sqrt{M}\in I\)đpcm

khó thế

kết quả là \(\sqrt{\left(2a+1\right)^2}+\left(2a+3\right)^2\varepsilon I\)đúng ko