Căn bậc hai. Căn bậc ba - Hỏi đáp

Bài đầu : \(\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}\)

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

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

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

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

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

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

Bài cuối : \(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

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

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

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

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

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

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

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

ĐKXĐ:x \(\ge\)0
\(\sqrt{25x}-\sqrt{16x}=9\)
\(\Leftrightarrow\)\(5\sqrt{x}-4\sqrt{x}=9\)
\(\Leftrightarrow\sqrt{x}=9\)
\(\Leftrightarrow x=3\)

mk viết nhầm nha ...Tìm giá trị của x để M\(^2\)<\(\dfrac{1}{4}\)

Từ \(a+b+c=1\Rightarrow2\left(a+b+c\right)=2\)

\(\Rightarrow\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=2\)

Và BĐT trên tương đương với

\(VT=\dfrac{c+ab}{a+b}+\dfrac{a+bc}{b+c}+\dfrac{b+ac}{a+c}\)

\(=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)

\(=\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)

Đặt \(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\)\(\left(x,y,z>0\right)\) thì ta có:

\(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge2\)\(\forall\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2\end{matrix}\right.\)

Đúng theo BĐT AM-GM

lập phương 2 vế ta có
x+1=8
\(\Leftrightarrow\)x=7

\(\sqrt[3]{x+1}=2\Leftrightarrow\left(\sqrt[3]{x+1}\right)^3=2^3\Leftrightarrow x+1=8\Leftrightarrow x=7\)

Vậy S={7}

Để so sánh \(\sqrt{3+\sqrt{20}}\) và \(\sqrt{5+\sqrt{5}}\)

Thì ta chỉ cần so sánh \(3+\sqrt{20}\) và \(5+\sqrt{5}\)

Hay ta chỉ cần so sánh \(\sqrt{20}\) và 2+\(\sqrt{5}\)

Mà \(\sqrt{20}=2\sqrt{5}>\sqrt{4}+\sqrt{5}=2+\sqrt{5}\)

\(\Rightarrow3+\sqrt{20}>5+\sqrt{5}\)

\(\Rightarrow\sqrt{3+\sqrt{20}}>\sqrt{5+\sqrt{5}}\)

\(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)

\(pt\Leftrightarrow\dfrac{x^2+4x+7}{x+4}=\sqrt{x^2+7}\)

\(\Leftrightarrow\dfrac{x^2+4x+7}{x+4}-4=\sqrt{x^2+7}-4\)

\(\Leftrightarrow\dfrac{x^2-9}{x+4}=\dfrac{x^2+7-16}{\sqrt{x^2+7}+4}\)

\(\Leftrightarrow\dfrac{x^2-9}{x+4}-\dfrac{x^2-9}{\sqrt{x^2+7}+4}=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(\dfrac{1}{x+4}-\dfrac{1}{\sqrt{x^2+7}+4}\right)=0\)

Xét pt \(\dfrac{1}{x+4}-\dfrac{1}{\sqrt{x^2+7}+4}=0\Leftrightarrow\dfrac{1}{x+4}=\dfrac{1}{\sqrt{x^2+7}+4}\)

\(\Leftrightarrow x+4=\sqrt{x^2+7}+4\Leftrightarrow x=\sqrt{x^2+7}\)

\(\Leftrightarrow x^2=x^2+7\Leftrightarrow0=7\) (vô nghiệm)

Nên \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)

\(\Leftrightarrow x^4+16x^2+49+8x^3+14x^2+56x=\left(x^2+8x+16\right)\left(x^2+7\right)\)

\(\Leftrightarrow x^4+8x^3+30x^2+56x+49=x^4+8x^3+23x^2+56x+112\)

\(\Leftrightarrow30x^2+49-23x^2-112=0\)

\(\Leftrightarrow7x^2-63=0\)

\(\Leftrightarrow7\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

a) Xét \(\Delta\)ABC vuông tại A

có: \(tanB=\dfrac{AC}{AB}=\dfrac{12}{9}=\dfrac{4}{3}\)

=> \(\widehat{B}=53^o\)

Lại có: \(AB^2+AC^2=BC^2\)

=> \(BC=\sqrt{9^2+12^2}=15\left(cm\right)\)

b) Xét \(\Delta\)AHB vuông tại H: HE \(\perp\) AB

có: \(AH^2=AB\cdot AE\)

Xét \(\Delta\)AHC vuông tại H: HF \(\perp\) AC

có: \(AH^2=AF\cdot AC\)

do đó: \(AB\cdot AE=AF\cdot AC\)

Cho hỏi chút... Đề đã cho AB=9cm... Sao câu a lại bảo tính độ dài AB hửm