\(\left|4-x\right|+2x=0\)

`TH1:x<=4`

`(4-x)+2x=0`

`=>4-x+2x=0`

`=>x+4=0`

`=>x=-4(tm)` 

`TH2:x>4`

`-(4-x)+2x=0`

`=>-4+x+2x=0`

`=>3x-4=0`

`=>3x=4`

`=>x=4/3` (ktm)

Vậy: ... 

xy-x=3y+10

=>x(y-1)-3y=10

=>x(y-1)-3y+3=13

=>(x-3)(y-1)=13

=>\(\left(x-3;y-1\right)\in\left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(4;14\right);\left(16;2\right);\left(2;-12\right);\left(-10;0\right)\right\}\)

\(C=\dfrac{5}{2\cdot7}+\dfrac{16}{7\cdot9}-\dfrac{2}{9\cdot11}-\dfrac{29}{1\cdot11}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{29}{11}\)

\(=\dfrac{1}{2}-\dfrac{28}{11}=\dfrac{11-56}{22}=\dfrac{-45}{22}< \dfrac{1}{3}\)

67

(x+4)(y-5)=2

=>(x+4;y-5)\(\in\){(1;2);(2;1);(-1;-2);(-2;-1)}

=>(x;y)\(\in\){(-3;7);(-2;6);(-5;3);(-6;4)}

Xét ΔABC có

BE,CF là các đường cao

BE cắt CF tại H

Do đó: H là trực tâm của ΔABC

=>AH\(\perp\)BC

a: 25-(5-x)=-7

=>5-x=25+7=32

=>x=5-32=-27

b: \(\dfrac{1}{2}+\dfrac{2}{3}:\left(x-1\right)=\dfrac{3}{4}\)

=>\(\dfrac{2}{3}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)

=>\(x-1=\dfrac{2}{3}:\dfrac{1}{4}=\dfrac{2}{3}\cdot4=\dfrac{8}{3}\)

=>\(x=\dfrac{8}{3}+1=\dfrac{11}{3}\)

d: \(\left(2x+1\right)\left(x-\dfrac{1}{7}\right)=0\)

=>\(\left[{}\begin{matrix}2x+1=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{7}\end{matrix}\right.\)

c: \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

=>\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}=\dfrac{7}{21}-\dfrac{9}{21}=-\dfrac{2}{21}\)

=>\(x=-\dfrac{2}{21}:\dfrac{4}{7}=-\dfrac{2}{21}\cdot\dfrac{7}{4}=\dfrac{-1}{6}\)

Thay a=1/3;b=3/5 vào A, ta được:

\(A=3\cdot\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{3}{5}+\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot\dfrac{3}{5}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{4}{5}+\dfrac{1}{10}=\dfrac{9}{10}\)

a: Vì \(\widehat{yCB}=\widehat{yDA}\left(=76^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên BC//AD
b: z là ở đâu bạn ơi?

`N = x^2 +3|y-2| -1`

Ta có: `{(x^2 >= 0 ),(|y-2| >=0):}`

`=> {(x^2 >= 0 ),(3|y-2| >=0):}`

`=> x^2 +3|y-2| >= 0`

`=> x^2 +3|y-2| -1 >=- 1`

Hay `N >= -1`

Dấu = xảy ra khi: 

`{(x^2 = 0 ),(|y-2| =0):}`

`<=> {(x = 0 ),(y-2 =0):}`

`<=> {(x = 0 ),(y=2):}`

Vậy `N_(min) = -1 <=> {(x = 0 ),(y=2):}`

-------------------------------------------------

`K = ( x+2)^2+( y-1/5)^2 -8`

Ta có: `{(( x+2)^2 >=0),(( y-1/5)^2 >=0):}`

`=> ( x+2)^2+( y-1/5)^2 >= 0`

`=>  ( x+2)^2+( y-1/5)^2 -8 >=- 8`

Hay `K >= -8`

Dấu = xảy ra khi: 

`{(( x+2)^2 =0),(( y-1/5)^2 =0):}`

`<=> {( x+2 =0),( y-1/5 =0):}`

`<=> {( x=-2),( y=1/5):}`

Vậy `K_(min) = -8 <=> {( x=-2),( y=1/5):}`

Bài 6:

a) Ta có:

\(\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>2\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>A=2\sqrt{x+7}-5\ge-5\forall\left(x\ge-7\right)\)

Dấu "=" xảy ra: `x+7=0`

`<=>x=-7` 

b) Ta có:

\(\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>A=-12+\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\)

Dấu "=" xảy ra: `x-8<=>x=8` 

Bài 7:

a: ĐKXĐ: x>=0

\(-\dfrac{1}{4}\sqrt{x}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x}+4< =4\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

b: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-2\end{matrix}\right.\)

\(\sqrt{x^2-4}>=0\forall x\) thỏa mãn DKXĐ
=>\(-\dfrac{1}{4}\sqrt{x^2-4}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x^2-4}+1< =1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(x^2-4=0\)

=>\(x^2=4\)

=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)