Bài 4: Phương trình tích

`(4x-1)(x-3)-2x+6=0`

`(4x-1)(x-3)-2(x-3)=0`

`<=> (x-3).(4x-3)=0`

`<=>` \(\left[{}\begin{matrix}x-3=0\\4x-3=0\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=0+3=3\\4x=0+3=3\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=3\\x=\dfrac{3}{4}\end{matrix}\right.\)

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

=>(4x-1)(x-3)-2(x-3)=0

=>(x-3)(4x-3)=0

=>x=3 hoặc x=3/4

Cần gấp ^^

\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)

=>x+100=0

=>x=-100

a: =>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+8}{92}+1\right)+\left(\dfrac{x+3}{97}+1\right)=0\)

=>x+100=0

=>x=-100

b: =>(x-11/111+1)+(x-12/112+1)=(x-23/123+1)+(x-24/124+1)

=>x+100=0

=>x=-100

Câu 2b) Ta có:

\(VT=\left[\left(x-3\right)^2+2\right]\left[\left(y+1\right)^2+3\right]\ge2.3=6\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

\(VP=-\left(z^2-4z+4\right)+6=-\left(z-2\right)^2+6\le6\)

Dấu "=" xảy ra \(\Leftrightarrow...\Leftrightarrow z=2\)

Vì \(VT=VP\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\\z=2\end{matrix}\right.\)

Câu 5: (đây là 1 dạng toán tư duy, nên bài này ở dạng khó).

Gọi \(h_A,h_B,h_C,h_D\) lần lượt là các đường cao hạ từ D,A,B,C đến các cạnh OA,OB,OC,OD.

Ta có: \(S=S_{OAD}+S_{OAB}+S_{OBC}+S_{OCD}\)

\(=\dfrac{1}{2}\left(OA.h_D+OB.h_A+OC.h_B+OD.h_C\right)\)

\(\Rightarrow OA.h_D+OB.h_A+OC.h_B+OD.h_C=OA^2+OB^2+OC^2+OD^2\)

Mặt khác, theo quan hệ giữa đg xiên và đg vuông góc, ta có:

\(h_D\le OD;h_A\le OA;h_B\le OB;h_C\le OC\)

\(\Rightarrow OA.OD+OB.OA+OC.OB+OD.OC\ge OA^2+OB^2+OC^2+OD^2\)(1)

Ta c/m BĐT: \(OA.OD+OB.OA+OC.OB+OD.OC\le OA^2+OB^2+OC^2+OD^2\left(2\right)\)

\(\Leftrightarrow2OA.OD+2OB.OA+2OC.OB+2OD.OC\le2OA^2+2OB^2+2OC^2+2OD^2\)

\(\Leftrightarrow0\le\left(OA-OD\right)^2+\left(OB^2-OA^2\right)+\left(OC^2-OB^2\right)+\left(OD^2-OC^2\right)\left(đúng\right)\)

Từ (1), (2) suy ra: 

\(OA.OD+OB.OA+OC.OB+OD.OC=OA^2+OB^2+OC^2+OD^2=OA.h_D+OB.h_A+OC.h_B+OD.h_C\)

Điều này chỉ xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}OA\perp OD,OB\perp OA,OC\perp OB\\OA=OB=OC=OD\end{matrix}\right.\)

\(\Leftrightarrow\)AC⊥BD tại O, \(OA=OB=OC=OD\)

\(\Leftrightarrow\)ABCD là hình vuông (có tâm O).

Câu 3:

b) \(a,b,c>0\)

 \(A=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Áp dụng BĐT AM-GM ta có:

\(A\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=\dfrac{8abc}{abc}=8\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Mà theo đề bài: \(A=8\)

\(\Rightarrow a=b=c\)

Vậy △ABC đều.

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

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

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

\(a,x^2+3x-10=0\\ =>x^2+5x-2x-10=0\\ =>x\left(x+5\right)-2\left(x+5\right)=0\\ =>\left(x-2\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ b,2x^2-3x-2=0\\ =>2x^2-x+4x-2=0\\ =>x\left(2x-1\right)+2\left(2x-1\right)=0\\ =>\left(x+2\right)\left(2x-1\right)=0\\ =>\left[{}\begin{matrix}x+2=0\\2x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)

1: =>(x+5)(x-2)=0

=>x=2 hoặc x=-5

2:=>2x^2-4x+x-2=0

=>(x-2)(2x+1)=0

=>x=2 hoặc x=-1/2

\(1,x^2+3x-10=0\)

\(\Leftrightarrow x^2-2x+5x-10=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

\(2,2x^2-3x-2=0\)

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

b: =>(x+9)(x-3)(x+21)=0

=>x+9=0; x-3=0; x+21=0

=>x=-9; x=3;x=-21

g: =>(x+2)(x+3)=0

=>x=-2 hoặc x=-3

d: =>x^2+4x+4=0

=>(x+2)^2=0

=>x+2=0

=>x=-2

D

Giải : 

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

\(2,2x^3+5x^2-3x=0\\ \Leftrightarrow2x^3+6x^2-x^2-3x=0\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2x^2-x\right)\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(3,6x^2-17x+12=0\\ \Leftrightarrow6x^2-8x-9x+12=0\\ \Leftrightarrow2x\left(3x-4\right)-3\left(3x-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(3x-4\right)=0\\\Leftrightarrow \left[{}\begin{matrix}2x-3=0\\3x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

1)

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

<=> x^2 +2x+1=4x^2 -8x+4`

`<=> 4x^2 -x^2 -8x -2x +4 -1 =0`

`<=> 3x^2 -10x +3=0`

`<=> 3x^2 -9x-x+3=0`

`<=> 3x(x-3)-(x-3)=0`

`<=> (x-3)(3x-1)=0`

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

2)

`2x^3 +5x^2 -3x=0`

`<=> x(2x^2 +5x-3)=0`

`<=> x(2x^2 +6x-x-3)=0`

`<=> x[2x(x+3)-(x+3)]=0`

`<=> x(x+3)(2x-1)=0`

\(< =>\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

3)

`6x^2 -17x+12=0`

`<=> 6x^2 -9x-8x+12=0`

`<=> 3x(2x-3)-4(2x-3)=0`

`<=> (2x-3)(3x-4)=0`

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

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

\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\)

\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)

\(\Leftrightarrow10x-3x^2-3=0\)

\(\Leftrightarrow-3x^2+10x-3=0\)

\(\Leftrightarrow-3x^2+9x+x-3=0\)

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

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

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

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{3;\dfrac{-1}{3}\right\}\)

\(2)2x^3+5x^2-3x=0\)

\(\Leftrightarrow2x^3+6x^2-x^2-3x=0\)

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

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-3;0;\dfrac{1}{2}\right\}\)

\(3)6x^2-17x+12=0\)

\(\Leftrightarrow6x^2-8x-9x+12=0\)

\(\Leftrightarrow\left(6x^2-8x\right)-\left(9x-12\right)=0\)

\(\Leftrightarrow2x\left(3x-4\right)-3\left(3x-4\right)=0\)

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

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

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{4}{3};\dfrac{3}{2}\right\}\)

`(3x-1)(x+1)=(9x^2-6x+1)`

`<=> 3x^2 + 3x -x-1 = 9x^2 -6x +1`

`<=> 3x^2 + 3x -x -9x^2 +6x = 1 +1`

`<=> -6x^2 -8x =2`

`<=> -6x^2 - 8x -2 =0`

`<=> -2(3x^2 +4x +1 )=0`

`<=> 3x^2 +4x +1 =0`

`=> Delta = 4^2 -4*3*1 = 4 >0`

`=> PT` có 2 nghiệm phân biệt 

\(\left[{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-4+\sqrt{4}}{2\cdot3}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-4-\sqrt{4}}{2\cdot3}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x_1=\dfrac{-4+2}{6}=\dfrac{-1}{3}\\x_2=\dfrac{-4-2}{6}=-1\end{matrix}\right.\)

Vậy `S={-1/3 ;-1}`