Câu hỏi của TRẦN HỮU ĐẠT - Toán lớp 9 - Học toán với OnlineMath

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

\(=\frac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}+1}\)\(+\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}+1}\)

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

\(=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)

\(P\in Z\Leftrightarrow1+\frac{3}{\sqrt{x}-1}\in Z\Rightarrow\frac{3}{\sqrt{x}-1}\in Z\)

\(\Rightarrow\sqrt{x}-1\inƯ_3\)

Mà \(Ư_3=\left\{\pm1;\pm3\right\}\)

\(Th1:\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(Th2:\sqrt{x}-1=-1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(Th3:\sqrt{x}-1=3\Rightarrow\sqrt{x}=4\Rightarrow x=16\)

\(Th4:\sqrt{x}-1=-3\Rightarrow\sqrt{x}=-2\Rightarrow x\in\varnothing\)

\(\Rightarrow x\in\left\{0;4;16\right\}\)

\(M=\frac{x+12}{\sqrt{x}-1}.\left(1\div\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)

\(=\frac{x-4+16}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{16}{\sqrt{x}+2}\)

\(=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\)

Áp dụng Bất đẳng thức Cô - Si cho hai số nguyên dương \(\sqrt{x}+2;\frac{16}{\sqrt{x}+2}\)ta có :

\(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}\)

\(\Rightarrow\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\ge2.\sqrt{16}=2.4=8\)

\(\Rightarrow\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\ge4\)

\(\Rightarrow M_{min}=4\Leftrightarrow\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\)

\(\Rightarrow\left(\sqrt{x}+2\right)^2=16\)

\(\Rightarrow\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(KL:M_{min}=4\Leftrightarrow x=4\)

AC=3cm

Sửa đề: n thuộc N*

n = 1 => mệnh đề đúng

Giả sử nó đúng đến n = k: \(7^k+3k-1⋮9\)

Cần chứng minh nó đúng với n = k + 1. \(7^{k+1}+3\left(k+1\right)-1⋮9\)

<=> \(7^k.7+3k+2=7\left(7^k+3k-1\right)-18k+9\)

\(=7\left(7^k+3k-1\right)-9\left(2k-1\right)⋮9\) (đúng)

P/s: Em có tính sai chỗ nào ko :>>

cứ thấy sai sai   xy+yz +ca =3

ta có \((\sqrt{a}-\sqrt{b})^2=a-2\sqrt{ab}+b\)

\(=a-b-2\sqrt{ab}+2b\)

\(=a-b-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)

VÌ a>b>0 NÊN \(\sqrt{a}-\sqrt{b}>0\)

suy ra : \(a-b-2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)< a-b\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< \left(\sqrt{a-b}\right)^2\)

VẬY \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\left(đ.p.c.m\right)\)

dien ak