Cho tam giác ABC có AC=2AB ,trên AC lấy D sao cho AB=2AD.Chứng minh:
a) Tam giác ABD đồng dạng với tam giác ACB
b) Góc ADB= Góc ABC
c) = AD.AC
giải hộ e vs ạ<3
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đề là \(\frac{5}{x+2}-3=0\) à
\(ĐKXĐ:x\ne-2\)
\(\frac{5-3x-6}{x+2}=0\)
\(\frac{-1-3x}{x+2}=0\)
\(-1-3x=0\)
\(x=-\frac{1}{3}\left(TM\right)\)
Giải thích các bước giải:
a,x+2x−2−x−2x+2=2(x+4)x2−4a,x+2x-2-x-2x+2=2(x+4)x2-4
⇒xx≠≠ −2,2-2,2
⇒x+2x−2−x−2x+2−2(x+4)x2−4=0x+2x-2-x-2x+2-2(x+4)x2-4=0
⇒(x+2)2−(x−2)2−(2x+8)(x−2)(x+2)=0(x+2)2-(x-2)2-(2x+8)(x-2)(x+2)=0
⇒6x−8(x−2)(x+2)=06x-8(x-2)(x+2)=0
⇒6x−8=06x-8=0
⇒x=43x=43
b,x−1x+6−xx−6=−7x2−36b,x-1x+6-xx-6=-7x2-36
⇒xx≠≠ −6,6-6,6
⇒x−1x+6−xx−6+−7x2−36=0x-1x+6-xx-6+-7x2-36=0
⇒(x−6)(x−1)−x(x+6)+7(x−6)(x+6)=0(x-6)(x-1)-x(x+6)+7(x-6)(x+6)=0
⇒−13x+13=0-13x+13=0
⇒x=1x=1
c,5x+2−5x−3+x2+16x2−x−6=0c,5x+2-5x-3+x2+16x2-x-6=0
⇒xx≠≠ −2,3-2,3
⇒5(x−3)−5(x+2)+x2+16(x+2)(x−3)=05(x-3)-5(x+2)+x2+16(x+2)(x-3)=0
⇒x2−9(x+2)(x−3)=0x2-9(x+2)(x-3)=0
⇒x+3x+2=0x+3x+2=0
⇒x+3=0x+3=0
⇒x=−3x=-3
d,x+3x+2−x−2x−3=x−3x2−x−6d,x+3x+2-x-2x-3=x-3x2-x-6
⇒xx≠≠ −2,3-2,3
⇒x+3x+2−x−2x−3=x−3(x+2)(x−3)x+3x+2-x-2x-3=x-3(x+2)(x-3)
⇒x+3x+2−x−2x−3−1x+2=0x+3x+2-x-2x-3-1x+2=0
⇒(x−3)(x+3)−(x+2)(x−2)−(x−3)(x+2)(x−3)=0(x-3)(x+3)-(x+2)(x-2)-(x-3)(x+2)(x-3)=0
⇒−2−x(x+2)(x−3)=0-2-x(x+2)(x-3)=0
⇒1x−3=01x-3=0
⇒xx ∈ ∅
a) \(A=\left(\frac{1}{a}+\frac{1}{a+1}\right):\frac{a}{a^2+a}\)
ĐKXĐ : a \(\ne0;a\ne-1\)
Khi đó A = \(\left(\frac{a+1+a^2}{a\left(a+1\right)}\right):\frac{a}{a\left(a+1\right)}=\frac{a^2+a+1}{a\left(a+1\right)}.\frac{a\left(a+1\right)}{a}=\frac{a^2+a+1}{a}\)
b) Khi (a - 5)(a + 1) = 0
<=> \(\orbr{\begin{cases}a-5=0\\a+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=5\left(\text{tmđk}\right)\\a=-1\left(\text{loại}\right)\end{cases}}\)
Thay a = 5 vào M ta được M = \(\frac{5^2+5+1}{5}=\frac{31}{5}\)
Vậy a = 5 thì M = 31/5
c) Ta có M = \(\frac{a^2+a+1}{a}=\frac{a^2-2a+1+3a}{a}=\frac{\left(a-1\right)^2}{a}+3\ge3\)
=> Min M = 3
Dấu "=" xảy ra <=> a - 1= 0 <=> a = 1 (tm)
Vậy Min M = 3 <=> a = 1
| 2-4x | = 4x-2
<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)
<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)
<=>\(\orbr{\begin{cases}-2+4x-4x+2=0\\2-4x-4x+2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)
<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)
=> \(S=\left\{\frac{1}{2};\infty\right\}\)
2x-7> 3(x-1)
<=>2x-7>3x-3
<=>2x-3x>-3+7
<=>-x>4
<=>x<4
=>S={x/x<4}
1-2x<4(3x-2)
<=>1-2x<12x-8
<=>-2x-12x<-8-1
<=>-14x<-9
<=>x>\(\frac{9}{14}\)
=>S={\(\frac{9}{14}\)}
-3x+2|-4 -x|> 0
<=>\(\orbr{\begin{cases}-3x+2+4+x>0\\-3x+2-4x-x>0\end{cases}}\)
<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)
<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)
<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)
=>S={x/x<3;x/x<\(\frac{1}{4}\)}
4x-1|x-2|< 0
<=>\(\orbr{\begin{cases}4x-1-x+2< 0\\4x-1+x-2< 0\end{cases}}\)
<=>\(\orbr{\begin{cases}3x+1< 0\\3x-3< 0\end{cases}}\)
<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)
<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)
=>S={x/x<\(\frac{-1}{3}\);x/x<1}
x3 - x2 - 8x + 12 = x3 - 2x2 + x2 - 2x - 6x + 12 = x2(x - 2) + x(x - 2) - 6(x - 2) = (x2 + x - 6)(x - 2)
= (x2 + 3x - 2x - 6)(x - 2) = (x + 3)(x - 2)(x - 2) = (x + 3)(x - 2)2
a) = ( x4 - x2 ) + ( x3 - 1 ) = x2( x - 1 )( x + 1 ) + ( x - 1 )( x2 + x + 1 ) = ( x - 1 )( x3 + 2x2 + x + 1 )
b) = x3( x - 3 ) - ( x - 3 ) = ( x - 3 )( x3 - 1 ) = ( x - 3 )( x - 1 )( x2 + x + 1 )
c) = 3( x + y ) - ( x + y )2 = ( x + y )( 3 - x - y )
d) = 2( x - y ) - ( x - y )2 = ( x - y )( 2 + x + y )
e) = y( x2 - y2 - 2y - 1 ) = y[ x2 - ( y + 1 )2 ] = y( x - y - 1 )( x + y + 1 )
f) = xz( x2 + 2xy + y2 - 9 ) = xz[ ( x + y )2 - 32 ] = xz( x + y - 3 )( x + y + 3 )
Câu C) \(AB^{2=}AD.AC\)nha các bạn