\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)

a: =x^4+2x^2+5x^3+10x-2x^2-4

=(x^2+2)(x^2+5x-2)

b; =(x+y)^3+z^3-3xy(x+y)-3xyz

=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)

=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1

=(x^2+x+1)(x^6-x^5+x^3-x^2+1)

.......... điểm danh nè!!!!!!!!!!!

me chưa ngủ

1d) giải theo các bước giải phương trình bậc 2 bình thường : x1 = 5 , x2 = 2 .

Người ÂM PHỦ ^_<

ko hiểu?????.tự hỏi,tự tl

a+b=1=> (a+b)²=1 => a²+2ab+b²=1 (*)

ta lại có

(a-b)² ≥ 0

<=> a²-2ab+b² ≥ 0 (**)

cộng (*) với (**)

a²+a²-2ab+2ab+b²+b² ≥ 1

<=> 2(a²+b²) ≥ 1

<=> a²+b² ≥ \(\dfrac{1}{2}\) (đpcm)

Mình cảm ơn bạn nhé ^^

Để mình giúp nha

\(x^2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+7=0\)

ĐKXD: x\(\ne0\)

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)

Đặt \(a=x+\dfrac{1}{x}\) khi đó phương trình trở thành

\(a^2-\dfrac{9}{2}a+5=0\)

\(\Leftrightarrow\left(a\right)^2-2.a.\dfrac{9}{4}+\left(\dfrac{9}{4}\right)^2-\dfrac{81}{16}+5=0\)

\(\Leftrightarrow\left(a+\dfrac{9}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}a-\dfrac{9}{4}=\dfrac{1}{4}\\a-\dfrac{9}{4}=-\dfrac{1}{4}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2+1}{x}=\dfrac{5}{2}\\\dfrac{x^2+1}{x}=2\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x\right)^2-2.x.\dfrac{5}{4}+\left(\dfrac{5}{4}\right)^2-\dfrac{25}{16}+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}=0\\x=1\end{matrix}\right.\)

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

Vậy S=\(\left\{1;2;\dfrac{1}{2}\right\}\)

cảm ơn Otasaka Yu

Câu 1:

A B C D M N

Có: \(S_{ABCD}=36cm^2\Rightarrow BC^2=36\Rightarrow BC=6cm\left(Vi:BC>0\right)\)

Vì: ABCD là hình vuông(gt)

=> BC=DC=AD=AB=6(cm)

Mà: M,N lần lượt là trung điểm của BC,CD

=>\(BM=MC=DN=NC=\frac{BC}{2}=\frac{6}{2}=3\left(cm\right)\)

CÓ: \(S_{AMN}=S_{ABCD}-\left(S_{ADN}+S_{ABM}+S_{NMC}\right)\)

\(=36-\left(\frac{1}{2}\cdot AD\cdot DN+\frac{1}{2}\cdot AB\cdot BM+\frac{1}{2}\cdot MC\cdot NC\right)\)

\(=36-\left(\frac{1}{2}\cdot6\cdot3+\frac{1}{2}\cdot6\cdot3+\frac{1}{2}\cdot3\cdot3\right)=\frac{27}{2}\left(cm^2\right)\)

Câu 2:

Có: \(a^2-6b^2=-ab\)

\(\Leftrightarrow\left(a^2-2ab\right)+\left(3ab-6b^2\right)=0\)

\(\Leftrightarrow a\left(a-2b\right)+3b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a+3b\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}a=2b\left(tm\right)\\a=-3b\left(loai\right)\end{matrix}\right.\)

Với \(a=2b\) ta có:

\(M=\frac{2b\cdot2b}{2\left(2b\right)^2-3b^2}=\frac{4b^2}{8b^2-3b^2}=\frac{4b^2}{5b^2}=\frac{4}{5}\)

câu 1 thì dễ òi, bn thi đc bn điểm

Bài 4:

a: \(2x^4+18x^2=0\)

=>\(2x^2\left(x^2+9\right)=0\)

=>\(x^2=0\) (Vì \(2\left(x^2+9\right)=2x^2+18\ge18>0\forall x\) )

=>x=0

b: (x-5)(x+5)-15x+75=0

=>(x-5)(x+5)-15(x-5)=0

=>(x-5)(x+5-15)=0

=>(x-5)(x-10)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=10\end{array}\right.\)

c: \(x^4=x^2\)

=>\(x^4-x^2=0\)

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

=>\(\left[\begin{array}{l}x^2=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=0\\ x^2=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

d: \(12x\left(6x-1\right)-24x^2=0\)

=>12x(6x-1-2x)=0

=>x(4x-1)=0

=>\(\left[\begin{array}{l}x=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac14\end{array}\right.\)

Bài 2:

a: 4x-16+3y(4-x)

=4(x-4)-3y(x-4)

=(x-4)(4-3y)

b: \(9y^2-6y+1=\left(3y\right)^2-2\cdot3y\cdot1+1^2=\left(3y-1\right)^2\)

c: \(25x^2-4=\left(5x\right)^2-2^2=\left(5x-2\right)\left(5x+2\right)\)

d: \(x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)

e: \(8x^3+36x^2+54x+27\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=\left(2x+3\right)^3\)

f: \(\left(2x-5\right)^2-\left(2x+y\right)^2\)

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

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

g: \(\left(2x-y\right)^3+\left(2x+y\right)^3\)

\(=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3\)

\(=16x^3+12xy^2=4x\left(4x^2+3y^2\right)\)

Câu 1:

a: \(6x^2-72x=0\)

=>\(6\left(x^2-12x\right)=0\)

=>\(x^2-12x=0\)

=>x(x-12)=0

=>\(\left[\begin{array}{l}x=0\\ x-12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=12\end{array}\right.\)

b: \(-2x^4+16x=0\)

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

=>\(x\left(x^3-8\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^3-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^3=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

c: \(\left(2x-1\right)^3-8x\left(x-3\right)\cdot\left(x+3\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x\cdot\left(x^2-9\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x^3+72x=-1\)

=>\(-12x^2+78x=0\)

=>-6x(2x-13)=0

=>x(2x-13)=0

=>\(\left[\begin{array}{l}x=0\\ 2x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{2}\end{array}\right.\)

d: \(x\left(x-5\right)-\left(x-3\right)^2=0\)

=>\(x^2-5x-\left(x^2-6x+9\right)=0\)

=>\(x^2-5x-x^2+6x-9=0\)

=>x-9=0

=>x=9

e: \(x\left(x-5\right)+3\left(x-5\right)=0\)

=>(x-5)(x+3)=0

=>\(\left[\begin{array}{l}x-5=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-3\end{array}\right.\)

f: 2x(x-8)-5(8-x)=0

=>2x(x-8)+5(x-8)=0

=>(x-8)(2x+5)=0

=>\(\left[\begin{array}{l}x-8=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-\frac52\end{array}\right.\)

g: \(30x-15x^2=0\)

=>15x(2-x)=0

=>x(2-x)=0

=>\(\left[\begin{array}{l}x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

h: \(-4x^3-12x=0\)

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

=>x=0

Tham khảo

đùa nhau à cái này làm sao pt dc

Gửi Thắng Nguyễn: Mình không biết tại sao lại ko phân tích được?