a: BD+DE=BE

CE+ED=CD
mà BD=CE

nên BE=CD

Xét ΔABE và ΔACD có

AB=AC

\(\hat{ABE}=\hat{ACD}\)

BE=CD

Do đó: ΔABE=ΔACD

=>\(\hat{EAB}=\hat{DAC}\)

b: MD+DB=MB

ME+EC=MC

mà DB=EC và MB=MC

nên MD=ME

Xét ΔAMD và ΔAME có

AM chung

MD=ME

AD=AE

Do đó: ΔAMD=ΔAME

=>\(\hat{MAD}=\hat{MAE}\)

=>AM là phân giác của góc DAE

c: ΔADE cân tại A

=>\(\hat{ADE}=\hat{AED}=\frac{180^0-\hat{DAE}}{2}=\frac{180^0-60^0}{2}=60^0\)

Bài 1 : 

Thay x = 2 ; y = -1/2 ta được 

\(B=-8+2.4\left(-\dfrac{1}{2}\right)-4.2.\left(\dfrac{1}{4}\right)+2\left(-\dfrac{1}{2}\right)-3\)

\(=-8-4-2-1-3=-18\)

a) Ta có: \(\left(2x-3\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};-\dfrac{4}{3}\right\}\)

b) Ta có: \(x^3-3x^2+3x-1=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2-2x+1-x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)

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

Vậy: S={0;1;3}

c) Ta có: \(x^2+x=2x+2\)

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

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

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

Vậy: S={-1;2}

d) Ta có: \(\left(x-1\right)^2=2\left(x^2-1\right)\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-1-2x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)Vậy: S={1;-3}

e) Ta có: \(2\left(x+2\right)^2-x^3-8=0\)

\(\Leftrightarrow2\left(x+2\right)^2-\left(x^3+8\right)=0\)

\(\Leftrightarrow2\left(x+2\right)\cdot\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\cdot\left(-x^2+4x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\end{matrix}\right.\)

Vậy: S={0;-2;4}

a: \(P=-\left|5-x\right|+2019\le2019\forall x\)

Dấu '=' xảy ra khi x=5

a: P(x)=2x^3-2x^3+x^2+3x^2-4x^2-3x+5x+1=-3x+6

b: P(0)=-3*0+6=6

P(-1)=6+3=9

P(1/3)=-1+6=5

c: P(x)=0

=>-3x+6=0

=>-3x=-6

=>x=2

P(x)=1

=>-3x+6=1

=>-3x=-5

=>x=5/3

`7,`

`a, B+A=4x-2x^2+3`

`-> B=(4x-2x^2+3)-A`

`-> B=(4x-2x^2+3)-(x^2-2x+1)`

`B=4x-2x^2+3-x^2+2x-1`

`B=(-2x^2-x^2)+(4x+2x)+(3-1)`

`B=-3x^2+6x+2`

`b, C-A=-x+7`

`-> C=(-x+7)+A`

`-> C=(-x+7)+(x^2-2x+1)`

`-> C=-x+7+x^2-2x+1`

`C=x^2+(-x-2x)+(7+1)`

`C=x^2-3x+8`

`c,`

`A-D=x^2-2`

`-> D= A- (x^2-2)`

`-> D=(x^2-2x+1)-(x^2-2)`

`D=x^2-2x+1-x^2+2`

`D=(x^2-x^2)-2x+(1+2)`

`D=-2x+3`

`6,`

`a,`

`P+Q=4x-2x^2+3`

`-> Q=(4x-2x^2+3)-P`

`-> Q=(4x-2x^2+3)-(3x^2+x-1)`

`Q=4x-2x^2+3-3x^2-x+1`

`Q=(-2x^2-3x^2)+(4x-x)+(3+1)`

`Q=x^2+3x+4`

`b,`

`x^2-5x+2-P=H`

`-> H= (x^2-5x+2)-(3x^2+x-1)`

`H=x^2-5x+2-3x^2-x+1`

`H=(x^2-3x^2)+(-5x-x)+(2+1)`

`H=-4x^2-6x+3`

`c,`

`P-R=5x^2-3x-4`

`-> R= P- (5x^2-3x-4)`

`-> R=(3x^2+x-1)-(5x^2-3x-4)`

`R=3x^2+x-1-5x^2+3x+4`

`R=(3x^2-5x^2)+(x+3x)+(-1+4)`

`R=-2x^2+4x+3`

a: Xét (O) có

ΔADB nội tiếp

AB là đường kính

Do đó: ΔADB vuông tại D

=>AD⊥BC tại D

Xét tứ giác AHDC có \(\hat{AHC}=\hat{ADC}=90^0\)

nên AHDC là tứ giác nội tiếp

b: AHDC nội tiếp

=>\(\hat{AHD}+\hat{ACD}=180^0\)

mà \(\hat{AHD}+\hat{MHD}=180^0\) (hai góc kề bù)

nên \(\hat{MHD}=\hat{ACD}=\hat{ACB}\)

Xét ΔOAC vuông tại A có AH là đường cao

nên \(OH\cdot OC=OA^2\)

=>\(OH\cdot OC=OB^2\)

=>\(\frac{OH}{OB}=\frac{OB}{OC}\)

Xét ΔOHB và ΔOBC có

\(\frac{OH}{OB}=\frac{OB}{OC}\)

góc HOB chung

Do đó: ΔOHB~ΔOBC

=>\(\hat{OHB}=\hat{OBC}=\hat{ABC}\)

mà \(\hat{OHB}+\hat{MHB}=\hat{OHM}=90^0\) và \(\hat{ABC}+\hat{ACB}=90^0\) (ΔABC vuông tại A)

nên \(\hat{MHB}=\hat{ACB}\)

=>\(\hat{MHB}=\hat{DHM}\)

=>HM là phân giác của góc DHB

used to deliver

used to be

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used to serve

1. used to deliver

2. used to be 

3. used to go

4. used to drive

5. used to spend

6. used to believe

7.used to work

8. used to serve