Bài 3: Liên hệ giữa phép nhân và phép khai phương

Áp dụng bđt AM-GM cho 2 số không âm ta có:

\(\dfrac{1}{\sqrt{1.2006}}>\dfrac{1}{\dfrac{1+2006}{2}}=\dfrac{2}{2007}\)

TT: \(\dfrac{1}{\sqrt{2.2005}}>\dfrac{2}{2007}\)

...

\(\dfrac{1}{\sqrt{2006.1}}>\dfrac{2}{2007}\)

Cộng vế với vế ta được:

\(S>\dfrac{2}{2007}.2006\)

ko đc tag tên có đc lm ko

Ta có : \(A^2=\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{6}=5+2\sqrt{6}=5+\sqrt{2^2.6}=5+\sqrt{24}.\)

Lại có : \(B^2=\left(\sqrt{10}\right)^2=10\)

\(\sqrt{24}< \sqrt{25}=>\sqrt{24}< 5=>5+\sqrt{24}< 5+5=>5+\sqrt{24}< 10\)

Mà \(A^2< B^2=>A< B=>\sqrt{2}+\sqrt{3}< \sqrt{10}\)

chtt có mà: Câu hỏi của nguyễn thị oanh - Toán lớp 9 | Học trực tuyến

\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)

đk : \(x\ne4\) ; \(x\ge0\)

1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)

Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)

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

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

Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

Q = \(\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\) = \(\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

Q = \(\dfrac{3}{2+\sqrt{x}}\)

b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)

\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)

c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4

ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên

\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3

mà \(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3

\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)

vậy x = 1 nguyên thì Q nguyên

2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)

= \(4\sqrt{a}-5\sqrt{10a}\)

b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)

c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

= \(33-2\sqrt{22}-11+3\sqrt{22}\)

= \(22+\sqrt{22}\)

Từ giả thiết \(\Rightarrow\dfrac{a+b+c}{2}=ax+by+cz=ax+2a=a\left(x+2\right)\).

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{2a}{a+b+c}\left(1\right)\)

Tương tự:

\(\dfrac{1}{y+2}=\dfrac{2b}{a+b+c}\left(2\right)\)

\(\dfrac{1}{z+2}=\dfrac{2c}{a+b+c}\left(3\right)\)

Cộng (1),(2) và (3) vế theo vế ta có :

\(M=\dfrac{1}{x+2}+\dfrac{1}{y+2}+\dfrac{1}{z+2}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2.\)

Vậy M=2.

ĐKXĐ: x>=0; y>=1 ; z>=2.

câu 1:Từ giả thiết ta có:

\(2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\sqrt{y-1}+1+\left(z-2\right)-2\sqrt{z-2}+1=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x}=1;\sqrt{y-1}=1;\sqrt{z-2}=1\)

Vậy x=1;y=2;z=3.

Có gì ko hiểu bạn cứ bình luận phía dưới :)

a)\(pt\Leftrightarrow\sqrt{3x^2-6x+4}+\sqrt{2x^2-4x+6}+x^2-2x-2=0\)

\(\Leftrightarrow\sqrt{3x^2-6x+4}-1+\sqrt{2x^2-4x+6}-2+x^2-2x+1=0\)

\(\Leftrightarrow\dfrac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}+\dfrac{2x^2-4x+6-4}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}+\dfrac{2\left(x-1\right)^2}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1\right)=0\)

Dễ thấy: \(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1>0\)

\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

b)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=3-4x-2x^2\)

\(pt\Leftrightarrow\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}+2x^2+4x-3=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x-8=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x+2=0\)

\(\Leftrightarrow\dfrac{3x^2+6x+12-9}{\sqrt{3x^2+6x+12}+3}+\dfrac{5x^4-10x^2+9-4}{\sqrt{5x^4-10x^2+9}+2}+2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\right)=0\)

Dễ thấy: \(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2>0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)

Ta có:

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(ab+ac+b^2+bc\right)\left(c+a\right)}{abc}\)

\(=\dfrac{b^2\left(c+a\right)}{abc}=\dfrac{b\left(c+a\right)}{ac}=\dfrac{bc+ab}{ac}=\dfrac{-ac}{ac}=-1\)

Ta có: \(ab+bc+ca=0\) => \(bc+ca=-ab\)

Ta lại có: P = \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}\)

= \(\dfrac{\left(a+b\right)\left(ba+bc+ca+c^2\right)}{abc}\) = \(\dfrac{\left(a+b\right)c^2}{abc}\) ( ab + ac+ bc = 0)

= \(\dfrac{ac^2+bc^2}{abc}\) = \(\dfrac{c\left(ac+bc\right)}{abc}\) = \(\dfrac{c.\left(-ab\right)}{abc}\) = \(-1\)
P/s: Bn có thể lm bài này theo 3 cách!

Ta có: \(P=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)

\(=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\)

\(=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2z}{xyz}-\dfrac{2x}{xyz}-\dfrac{2y}{xyz}\)

\(=3-\dfrac{2\left(x+y+z\right)}{xyz}\)

\(=3-\dfrac{2xyz}{xyz}=3-2=1\)

Vậy P = 1

Đề ??

\(1=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\left(b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{a}{bc}\right)\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{bc+ac+ab+a^2}{abc}\right)=0\)

\(\dfrac{\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc}=0\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)

Xét 3 TH

=> P=0 ( đề bài BT ở giữa có 1 số mũ sai nha )

pt(1)=>a=90-b thay vào pt(2) giải