a) thay x = -3/2 vào pt được : \(\left(-\frac{3}{2}\right)^2-m.\left(-\frac{3}{2}\right)+m+1=0\Leftrightarrow m=-\frac{13}{10}\)

mà theo định lí Vi-et thì : x₁+x₂=m => x₁=m-x₂= -13/10+3/2=1/5 (giả sử x₂ = -3/2)

b) tương tự

ĐKXĐ: \(x\ge0;x\ne4.\)

\(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}.\)

\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}.\)

b) Để \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\Leftrightarrow\frac{4\sqrt{x}}{4\left(\sqrt{x}-2\right)}-\frac{5\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}=0\)

\(\Leftrightarrow\frac{4\sqrt{x}-5\sqrt{x}+10}{4\left(\sqrt{x}-2\right)}=0\Leftrightarrow-\sqrt{x}+10=0\)

\(\Leftrightarrow\sqrt{x}=10\Leftrightarrow x=100\left(tmđk\right).\)

Vậy để A=5/4 thì x=100

Tự tìm ĐK nha

a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

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

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

\(A=\frac{\sqrt{x}}{\sqrt{x}-2}\)

b) \(A=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\)

\(\Leftrightarrow4\sqrt{x}=5\left(\sqrt{x}-2\right)\)

\(\Leftrightarrow4\sqrt{x}=5\sqrt{x}-10\)

\(\Leftrightarrow\sqrt{x}=10\)

\(\Leftrightarrow x=100\)( thỏa mãn )

Vậy...

Giúp mình với TvT . Mình cảm ơn huhu

B1: \(\sqrt{x-9+2\sqrt{x-9}+1}\)=x-20

\(\sqrt{\left(\sqrt{x-9}+1\right)^2}\)=x-20

\(\sqrt{x-9}=x-21\)

bình phương lên giải pt bậc 2

câu hỏi tương tự

câu hỏi tương tự đã ai trả lời đâu

\(\hept{\begin{cases}mx+my=-3\\\left(1-m\right)x+y=0\end{cases}}\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-3}{m^2}\\y=\left(m-1\right).\frac{-3}{m^2}\end{cases}}\)

Để phương trình có nghiệm âm thì ta có

\(\hept{\begin{cases}\frac{-3}{m^2}< 0\\\frac{-3.\left(m-1\right)}{m^2}< 0\end{cases}}\Leftrightarrow m>1\)

Cảm ơn a ạ!! :))

Áp dụng BĐT Cô-si cho 2 số dương ta có:

\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\left(1\right)\)

\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\left(2\right)\)

\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\left(2\right)\)

Từ (1) ;(2) và (3) suy ra:

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=6\)

Vậy \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\).Dấu "=" xảy ra <=>\(\hept{\begin{cases}a+b+c=6abc\\\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}\end{cases}=>a=b=c=\frac{1}{\sqrt{2}}}\)

A = \(x-2\sqrt{xy}+3y-2\sqrt{x}+1\)

\(=\left(\frac{x}{3}-\frac{2\times\sqrt{3}\sqrt{xy}}{\sqrt{3}}+3y\right)+\left(\frac{2x}{3}-\frac{2\times\sqrt{2}\times\sqrt{3}\sqrt{x}}{\sqrt{2}\times\sqrt{3}}+\frac{3}{2}\right)-\frac{1}{2}\)

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

\(\ge-\frac{1}{2}\)