Bài 4:

a: Đặt 2x+10=0

=>2x=-10

=>x=-5

b: Đặt 4(x-1)+3x-5=0

=>4x-4+3x-5=0

=>7x=9

=>\(x=\dfrac{9}{7}\)

c: Đặt \(-1\dfrac{1}{3}x^2+x=0\)

=>\(\dfrac{4}{3}x^2-x=0\)

=>\(x\left(\dfrac{4}{3}x-1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\\dfrac{4}{3}x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\)

Bài 5:

a: Xét ΔBAD vuông tại A và ΔBMD vuông tại M có

BD chung

\(\widehat{ABD}=\widehat{MBD}\)

Do đó: ΔBAD=ΔBMD

b: ΔBAD=ΔBMD

=>BA=BM và DA=DM

ta có: BA=BM

=>B nằm trên đường trung trực của AM(1)

Ta có: DA=DM

=>D nằm trên đường trung trực của AM(2)

Từ (1),(2) suy ra BD là đường trung trực của AM

c: Xét ΔBKC có

KM,CA là các đường cao

KM cắt CA tại D

Do đó: D là trực tâm của ΔBKC

=>BD\(\perp\)KC tại N

Bài 2:

Độ dài của `1/3` quãng đường đầu là: 

`1/3*600=200` (km) 

Thời gian xe đi trên `1/3` quãng đường đầu là:

\(\dfrac{200}{x}\left(h\right)\)

Quãng đường còn lại là: `600 - 200 = 400`(km) 

Vận tốc của xe khi đi trên quãng đường còn lại: `x+10` (km/h) 

Thời gian xe đi trên quãng đường còn lại là:

\(\dfrac{400}{x+10}\left(h\right)\) 

Biểu thức thể hiện thời gian xe đi từ Hà Nội đến Quãng Ngãi là:

\(\dfrac{200}{x}+\dfrac{400}{x+10}=\dfrac{200\left(x+10\right)}{x\left(x+10\right)}+\dfrac{400x}{x\left(x+10\right)}=\dfrac{200x+2000+400x}{x\left(x+10\right)}=\dfrac{600x+2000}{x\left(x+10\right)}\)

+, Vì ABCD là hình vuông nên: \(\left\{{}\begin{matrix}AB=BC=CD=DA\\\widehat{ABC}=\widehat{BCD}=\widehat{CDA}=\widehat{DAB}=90^{\circ}\end{matrix}\right.\) (t/c)

Xét \(\triangle DEA\) và \(\triangle DFC\) có: \(\left\{{}\begin{matrix}AE=CF\left(gt\right)\\\widehat{DAE}=\widehat{DCF}=90^{\circ}\\AD=CD\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\triangle DEA=\triangle DFC\left(c.g.c\right)\Rightarrow\left\{{}\begin{matrix}DE=DF\left(\text{hai cạnh tương ứng}\right)\\\widehat{ADE}=\widehat{CDF}\left(\text{hai góc tương ứng}\right)\end{matrix}\right.\)

+, Lại có: \(\widehat{ADE}+\widehat{DCE}=\widehat{ADC}=90^{\circ}\)

\(\Rightarrow\widehat{CDF}+\widehat{DCE}=90^{\circ}\Rightarrow\widehat{EDF}=90^{\circ}\)

+, Xét tứ giác \(DEMF\) có: \(\left\{{}\begin{matrix}DM\cap EF=\left\{O\right\}\\O\text{ là trung điểm của }DM\left(gt\right)\\O\text{ là trung điểm của }EF\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow\) Tứ giác DEMF là hình bình hành

Mà: \(\widehat{EDF}=90^{\circ}\left(cmt\right)\Rightarrow DEMF\) là hình chữ nhật

Mặt khác: \(DE=DF\left(cmt\right)\)

Do đó: Tứ giác DEMF là hình vuông.

$\text{#}Toru$

1) Đặt: `x^2+x=t` 

Pt trở thành: 

`t^2+4t-12=0`

\(\Leftrightarrow t^2-2t+6t-12=0\\ \Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\\ \Leftrightarrow\left(t-2\right)\left(t+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=6\end{matrix}\right.\) 

Với `t=2` \(\Leftrightarrow x^2+x=2\Leftrightarrow x^2+x-2=0\Leftrightarrow x^2-x+2x-2=0\)

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

Với `t=6` \(\Leftrightarrow x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow x^2-2x+3x-6=0\)

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

Vậy: ... 

2) \(\left(x^2+2x+3\right)^2-9\left(x+1\right)^2-28=0\)

\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]^2-9\left(x+1\right)^2-28=0\\ \Leftrightarrow\left(x+1\right)^4+4\left(x+1\right)^2+4-9\left(x+1\right)^2-28=0\\ \Leftrightarrow\left(x+1\right)^4-5\left(x+1\right)^2-24=0\\ \Leftrightarrow\left(x+1\right)^4-8\left(x+1\right)^2+3\left(x+1\right)^2-24=0\\ \Leftrightarrow\left(x+1\right)^2\left[\left(x+1\right)^2-8\right]+3\left[\left(x+1\right)^2-8\right]=0\\ \Leftrightarrow\left[\left(x+1\right)^2+3\right]\left[\left(x+1\right)^2-8\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2+3\ge3>0\left(L\right)\\\left(x+1\right)^2-8=0\end{matrix}\right.\\ \Leftrightarrow\left(x+1\right)^2 =8\\ \Leftrightarrow\left[{}\begin{matrix}x+1=2\sqrt{2}\\x+1=-2\sqrt{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=-2\sqrt{2}-1\end{matrix}\right.\)

Vậy: ... 

a, Đặt x^2 + x = t 

\(t^2+4t-12=0\Leftrightarrow t=2;t=-6\)

\(\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\left(l\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b, \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+1\right)-28=0\)

Đặt x^2 + 2x + 1 = t ( t >= 0 ) 

\(\left(t+2\right)^2-9t-28=0\Leftrightarrow t^2-5t-24=0\Leftrightarrow t=8;t=-3\left(l\right)\)

\(\Rightarrow\left(x+1\right)^2=8\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=-2\sqrt{2}-1\end{matrix}\right.\)

c, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

Đặt x^2 - 4 = t 

\(t\left(t-6\right)=72\Leftrightarrow t^2-6t-72=0\Leftrightarrow t=12;t=-6\)

\(\left[{}\begin{matrix}x^2-4=12\\x^2-4=-6\left(l\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

d, \(x\left(x+1\right)\left(x^2+x+1\right)=42\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt x^2 + x = t 

\(t\left(t+1\right)=42\Leftrightarrow t^2+t-42=0\Leftrightarrow t=6;t=-5\)

\(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-5\left(l\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

1) 

\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2=\left(3x+6\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x=-13\\x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=1\end{matrix}\right.\)

Vậy: ... 

2) 

\(\left(x+2\right)=9\left(x^2+4x+4\right)\\ \Leftrightarrow\left(x+2\right)=9\left(x+2\right)^2\\ \Leftrightarrow9\left(x+2\right)^2-\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left[9\left(x+2\right)-1\right]=0\\ \Leftrightarrow\left(x+2\right)\left(9x+17\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\9x=-17\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{17}{9}\end{matrix}\right.\)

Vậy: ... 

3) 

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\ \Leftrightarrow\left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\ \Leftrightarrow\left(7x+23\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-23}{7}\\x=5\end{matrix}\right.\)

Vậy: ... 

4) 

\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\\ \Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\\\Leftrightarrow\left(5x^2-2x+10+3x^2+10x-8\right)\left(5x^2-2x+10-3x^2-10x+8\right)=0\\ \Leftrightarrow\left(8x^2+8x+2\right)\left(2x^2-12x+18\right)=0\\ \Leftrightarrow4\left(4x^2+4x+1\right)\left(x^2-6x+9\right)\\ \Leftrightarrow4\left(2x+1\right)^2\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=3\end{matrix}\right.\)

Vậy: ... 

\(A=12x-8y-4x^2-y^2+1\)

\(=\left(-4x^2+12x-9\right)+\left(-y^2-8y-16\right)+26\)

\(=-\left(4x^2-12x+9\right)-\left(y^2+8y+16\right)+26\)

\(=-\left(2x-3\right)^2-\left(y+4\right)^2+26\)

ta có: \(\left\{{}\begin{matrix}-\left(2x-3\right)^2\le0\\-\left(y+4\right)^2\le0\end{matrix}\right.\Rightarrow A=-\left(2x-3\right)^2-\left(y+4\right)^2+26\le0+0+26=26\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-4\end{matrix}\right.\)

Vậy: ... 

1, \(27x^2\left(x+3\right)-12x\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(27x^2-12x\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\Leftrightarrow x=0;x=-3;x=\dfrac{4}{9}\)

2, \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

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

3, \(\left(2x-1\right)^2=49\Leftrightarrow\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

4, \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\Leftrightarrow\left(5x-3\right)^2=\left(4x-7\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=4x-7\\5x-3=7-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)

1) \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\9x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{4}{9}\end{matrix}\right.\)

Vậy: ... 

2)  \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\) 

\(\Leftrightarrow\left(4x\right)^2-2\cdot4x\cdot1+1^2=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\)

\(\Leftrightarrow-13\left(4x-1\right)=0\)

\(\Leftrightarrow4x-1=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

Vậy: ... 

1) \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)=2\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5-2x-4\right)=0\)

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

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

Vậy: .... 

2) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)=-\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

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

Vậy: ... 

3) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)

Vậy: ... 

4) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-1\\5x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: ...

`#3107.101107`

`a + b + c = 0`

`=> (a + b + c)^3 = 0`

`=> a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc = 0`

`=> a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc + 3abc - 3abc = 0`

`=> a^3 + b^3 + c^3 + (3a^2b + 3ab^2 + 3abc) + (3b^2c + 3bc^2 + 3abc) + (3a^2c + 3ac^2 + 3abc) - 3abc = 0`

`=> a^3 + b^3 + c^3 + 3ab(a + b + c) + 3bc(b + c + a) + 3ac(a + c + b) = 3abc`

`=> a^3 + b^3 + c^3 + (3ab + 3bc + 3ac)(a + b + c) = 3abc`

Mà `a + b + c = 0`

`=> a^3 + b^3 + c^3 = 3abc` (đpcm).

a) \(x^2-36=0\)

\(\Leftrightarrow x^2-6^2=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\) 

Vậy: ... 

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy: ... 

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{6;-6\right\}\)

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2.x.5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)