Mk gợi ý qua cho bn r bn tự lm tiếp nhá

1)ĐK:\(x\in\left[-2;2\right]\)

Dễ thấy :y=0 không là nghiệm của hệ

Chia cả 2 vế của pt(2) cho \(y^3\)ta đc:

\(\left(5-x\right)\sqrt{2-x}=\frac{8}{y^3}+\frac{6}{y}\)\(\Leftrightarrow\left(2-x\right)\sqrt{2-x}+3\sqrt{2-x}=\left(\frac{2}{y}\right)^3+3\left(\frac{2}{y}\right)\)

Xét hàm số:\(f\left(t\right)=t^3+3t\)

\(f'\left(t\right)=3t^2+3>0\)\(\Rightarrow\)hàm số liên tục và đồng biến trên R

\(\Rightarrow\sqrt{2-x}=\frac{2}{y}\)\(\Rightarrow y=\frac{2}{\sqrt{2-x}}\)

Thay vào pt(1) ta đc:

\(6\sqrt{2+x}+8\sqrt{4-x^2}=20-6x+12\sqrt{2-x}\)

\(\Leftrightarrow x=\frac{6}{5}\Rightarrow y=\sqrt{5}\)(t/m)

KL:...

Giúp mình hệ 2 vs ạ :))

Các chị giỏi quá!👍👍

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Your younger brother goes to school by bike, doesn't he?

Some fresh flowers have been cut from the garden by my mother.

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a: góc ADH+góc AEH=180 độ

=>ADHE nội tiếp

góc BEC=góc BDC=90 độ

=>BEDC nội tiếp

b: góc EAH=90 độ-goc ABC

góc ECB=90 độ-góc ABC

=>góc EAH=góc ECB

c: góc xAC=góc ABC

=>góc xAC=góc ADE

=>xy//DE

lỗi

lỗi

Đường thẳng BC vuông góc AH nên nhận (1;-3) là 1 vtpt

Phương trình BC: \(1\left(x-2\right)-3\left(y+7\right)=0\Leftrightarrow x-3y-23=0\)

Do M thuộc CM nên tọa độ có dạng \(M\left(-2m-7;m\right)\)

M là trung điểm AB \(\Rightarrow A\left(-4m-16;2m+7\right)\)

Mà A thuộc AH nên:

\(3\left(-4m-16\right)+\left(2m+7\right)+11=0\Rightarrow m=-3\Rightarrow A\left(-4;1\right)\)

\(\Rightarrow\overrightarrow{AB}=\left(6;-8\right)\Rightarrow\) đường thẳng AB nhận (4;3) là 1 vtpt \(\Rightarrow\) pt AB là...

C là giao điểm BC và CM nên tọa độ thỏa mãn:

\(\left\{{}\begin{matrix}x+2y+7=0\\x-3y-23=0\end{matrix}\right.\) \(\Rightarrow C\left(5;-6\right)\Rightarrow\overrightarrow{BC}=...\Rightarrow\) phương trình BC

\(4,=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\\ =\dfrac{3\left(3-\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{6}}=\dfrac{\left(9-3\sqrt{2}-3\sqrt{3}\right)\left(\sqrt{6}-2\right)}{2}\\ =\dfrac{9\sqrt{6}-18-6\sqrt{3}+6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{2}\\ =\dfrac{9\sqrt{6}-3\sqrt{2}-18}{2}\)

\(7,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}+1-2-\sqrt{3}=1+\sqrt{2}\)

\(10,\dfrac{1}{\sqrt{a}+\sqrt{a+2}}=\dfrac{\sqrt{a}-\sqrt{a+2}}{a-a-2}=\dfrac{\sqrt{a-2}-\sqrt{a}}{2}\)

Do đó \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{47}+\sqrt{49}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{49}-\sqrt{47}}{2}=\dfrac{-1+\sqrt{49}}{2}=\dfrac{7-1}{2}=3\)

10, \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{17}+\sqrt{19}}=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{17}-\sqrt{19}}{\left(\sqrt{17}+\sqrt{19}\right)\left(\sqrt{17}-\sqrt{19}\right)}=\dfrac{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+...+\sqrt{17}-\sqrt{19}}{-2}=-\dfrac{1-\sqrt{19}}{2}\)

Để pt có 2 nghiệm dương:

\(\left\{{}\begin{matrix}\Delta'=\left(m-3\right)^2-\left(m-1\right)\ge0\\x_1+x_2=-2\left(m-3\right)>0\\x_1x_2=m-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-7m+10\ge0\\m< 3\\m>1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\ge5\\m\le2\end{matrix}\right.\\m< 3\\m>1\end{matrix}\right.\)

\(\Rightarrow1< m\le2\)

ĐKXĐ: \(x\ge\frac{1}{3}\)

\(x^2+5x=x\sqrt{3x-1}+\left(x+1\right)\sqrt{5x}\)

\(\Leftrightarrow2x^2+10x-2x\sqrt{3x-1}-2\left(x+1\right)\sqrt{5x}=0\)

\(\Leftrightarrow\left(x^2-2x\sqrt{3x-1}+3x-1\right)+\left[\left(x+1\right)^2-2\left(x+1\right)\sqrt{5x}+5x\right]=0\)\(\Leftrightarrow\left(x-\sqrt{3x-1}\right)^2+\left(x+1-\sqrt{5x}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\sqrt{3x-1}=0\\x+1-\sqrt{5x}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3x-1}\\x+1=\sqrt{5x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=3x-1\\\left(x+1\right)^2=5x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3x+1=0\\x^2-3x+1=0\end{matrix}\right.\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\left(tm\right)\)

`(15-x)+(x-12)=7-(-5+x)`

`=>15-x+x-12=7+5-x`

`=>3=12-x`

`=>x=12-3`

`=>x=9`

Vậy `x=9`

(15-x)+(x-12) = 7-(-5+x)

<=>15-x+x-12=7+5-x

<=>3=12-x

<=>x=12-3=9

(15-x)+(x-12) = 7-(-5+x)

<=> 15-x +x -12 = 7 +5 -x 

<=> x = 12 +12 -15

<=> x =9