a) Xét hpt : \(\hept{\begin{cases}x+my=1\\mx-3my=2m+3\end{cases}}\)

Tại m = -3 ta có :

\(\hept{\begin{cases}x-3y=1\\-3x+3.3y=-2.3+3\end{cases}}\)

<=> \(\hept{\begin{cases}x-3y=1\\-3x+9y=-3\end{cases}}\)

<=> \(\hept{\begin{cases}x-3y=1\\-x+3y=-1\end{cases}}\)

<=>\(\hept{\begin{cases}x-3y=1\\x-3y=1\end{cases}}\)

Do đó hpt có vô số nghiệm với m = -3

b) Xét hpt : \(\hept{\begin{cases}x+my=1\\mx-3ym=2m+3\end{cases}}\)

<=> \(\hept{\begin{cases}x=1-my\\m\left(1-my\right)-3ym=2m+3\end{cases}}\)

<=> \(\hept{\begin{cases}x=1-my\\m-m^2y-3my=2m+3\end{cases}}\)

<=> \(\hept{\begin{cases}x=1-my\\\left(m^2+3m\right)y=m-2m-3\end{cases}}\)

<=> \(\hept{\begin{cases}x=1-my\\\left(m^2+3m\right)y=-m-3\end{cases}}\)

Ta có : Hpt có nghiệm duy nhất

<=> Pt trên có nghiệm duy nhất

<=> m² + 3m khác 0

<=> m(m + 3) khác 0

<=> m khác 0 và m khác -3

=> Ta có :

\(\hept{\begin{cases}x=1-my\\m\left(m+3\right)y=-3-m\end{cases}}\)

<=> \(\hept{\begin{cases}y=\frac{-\left(m+3\right)}{m\left(m+3\right)}\\x=1-my\end{cases}}\)

<=> \(\hept{\begin{cases}x=2\\y=\frac{-1}{m}\end{cases}}\)

<=> \(\hept{\begin{cases}m\left(m+3\right)=0\\-\left(m+3\right)=0\end{cases}}\)

<=>\(\hept{\begin{cases}m=0orm=-3\\m=-3\end{cases}}\)

<=> m = -3

<=> m(m+3) = 0 và m(m + 3) khác 0

<=> m = 0 haowcj m = -3 và m khác -3

<=> m = 0

Vậy

ĐKXĐ : \(x\ge1;y\ge\frac{1}{2}\)

\(\hept{\begin{cases}\sqrt{x-1}+\sqrt{2y-1}=\sqrt{xy}\\\sqrt{x-1}-\sqrt{2y-1}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{2y-1}=\sqrt{xy}-1\\\sqrt{x-1}=\sqrt{2y-1}+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{2y-1}=\sqrt{xy}-1\\x-1=2y-1+2\sqrt{2y-1}+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2y-1=\sqrt{xy}-1\\\sqrt{x-1}+\sqrt{2y-1}=\sqrt{xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-\sqrt{xy}-2y=0\\\sqrt{x-1}+\sqrt{2y-1}=\sqrt{xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-2\sqrt{y}\right)=0\\\sqrt{x-1}+\sqrt{2y-1}=\sqrt{xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\sqrt{y}\left(tm\right)\\\sqrt{x-1}+\sqrt{2y-1}=\sqrt{xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4y\\\sqrt{4y-1}+\sqrt{2y-1}=\sqrt{4y^2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4y\\6y-2+2\sqrt{\left(4y-1\right)\left(2y-1\right)}=4y^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4y\\\sqrt{\left(4y-1\right)\left(2y-1\right)}=2y^2-3y+1=\left(2y-1\right)\left(y-1\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4y\\\left(2y-1\right)y^2\left(2y-5\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4y\\y=\frac{1}{2};y=\frac{5}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2;y=\frac{1}{2}\\x=10;y=\frac{5}{2}\end{cases}}\)

Vậy ....

Bình phương chuyển vế thành phương trình bậc 4 giải ra ta được x = -3 ; x = 2 

đặt ẩn phụ nhanh hơn đấy

duyhung723 nè

Ta có : \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)

\(\Rightarrow a^2c+b^2c-ab^2-ac^2=0\)

\(\Rightarrow a\left(ac-b^2\right)-c\left(ac-b^2\right)=0\)

\(\Rightarrow\left(a-c\right)\left(ac-b^2\right)=0\)

\(\Rightarrow ac=b^2\) ( do \(a\ne c\) )

\(\Rightarrow\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{c}{b}=\frac{b}{a}=\frac{a}{c}\)

\(\Rightarrow\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)

Áp dụng Cô si cho 2 số dương ta đc:

\(2\sqrt{4a\left(3a+b\right)}\le4a+\left(3a+b\right)=7a+b\)

Tương tự: \(2\sqrt{4b\left(3b+a\right)}\le4b+\left(3b+a\right)=7b+a\)

\(\Rightarrow2\sqrt{4a\left(3a+b\right)}+2\sqrt{4b\left(3b+a\right)}\le8\left(a+b\right)\)

\(\Leftrightarrow\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\le2\left(a+b\right)\)

\(\Leftrightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4a=3a+b\\4b=3b+a\\a,b>0\end{cases}}\Leftrightarrow a=b>0\)

Giải HPT:

\(\hept{\begin{cases}x+y-z=c\\y+z-x=a\\z+x-y=b\end{cases}\Leftrightarrow\hept{\begin{cases}2y=c+a\\2z=a+b\\2x=b+c\end{cases}\Leftrightarrow}}\hept{\begin{cases}y=\frac{c+a}{2}\\x=\frac{a+b}{2}\\x=\frac{b+c}{2}\end{cases}}\)

1 ) Áp dụng BĐT Cauchy : 

\(2\sqrt{a\left(3a+b\right)}=\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}\)

Tương tự \(2\sqrt{b\left(3b+a\right)}\le\frac{4b+3b+a}{2}\)

\(\Rightarrow2\left(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\right)\le\frac{8a+8b}{2}=4\left(a+b\right)\)

\(\Rightarrow\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\le2\left(a+b\right)\)

\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b>0\)

\(2x^2+2y^2=5xy\Leftrightarrow2x^2+2y^2-5xy=0\)

\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{y}{2}\\x=2y\end{cases}}\)

Mặt khác : x > y > 0 \(\Rightarrow x=2y\) 

Ta có : \(E=\frac{x+y}{x-y}=\frac{2y+y}{2y-y}=\frac{3y}{y}=3\)

a) Dễ tự làm đi

b) Xét 1 + a² = ab + bc + ca + a² 

                      = b(c + a) + a(c + a)

                      = (c + a)(b + a)

Cmtt ta có : 1 + b² = (c + b)(a + b)

                    1 + c² = (b+c)( a + c)

Do đó : A = \(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+b\right)\left(b+a\right)\left(c+a\right)\left(a+c\right)\left(b+c\right)}\)= 1

Xét a² + 2bc - 1 = a² + 2bc - ab - bc - ca

                           = a² - ab + bc - ca

                           = a(a-b) - c(a-b)

                           = (a-b)(a-c)

Cmtt ta cũng có : b² + 2ac - 1 = (b-c)(b-a)

                             c² + 2ab - 1 = (c-a)(c-b)

Do đó : \(B=\frac{\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ba-1\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)

                   \(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(b-a\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)

                     = -1

\(2\left(\sqrt{\frac{x^2+x+1}{x+4}}-1\right)+x^2-3=\frac{2}{\sqrt{x^2+1}}-1\)

\(\Leftrightarrow2\frac{\frac{x^2+x+1}{x+4}-1}{\sqrt{\frac{x^2+x+1}{x+4}}+1}+x^2-3=\frac{4-\left(x^2+1\right)}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\frac{2\left(x^2-3\right)}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+x^2-3=\frac{3-x^2}{\left(2\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\left(x^2-3\right)\left(\frac{2}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+1+\frac{1}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\right)=0\)

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(Cũng không chắc _-_ )

bạn làm đúng rồi đấy, mình đăng cho vuii thôi :)))