Gọi K là trung điểm của HA

Xét ΔHAD có

K,F lần lượt là trung điểm của HA,HD

=>KF là đường trung bình của ΔHAD

=>KF//AD và KF=AD/2

=>KF⊥BA

Xét ΔBAF có

FK,AH là các đường cao

FK cắt AH tại K

Do đó: K là trực tâm của ΔBAF

=>BK⊥AF

Ta có: \(KF=\frac{AD}{2}\)

\(BC=\frac{AD}{2}\)

Do đó: KF=BC

Ta có: KF//AD

AD//BC

Do đó: KF//BC

Xét tứ giác BCFK có

BC//FK

BC=FK

Do đó: BCFK là hình bình hành

=>BK//CF

mà BK⊥AF

nên FA⊥FC

=>\(\hat{CFA}=90^0\)

x²-4x=0

<=> x(x-4)=0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

Vậy x=0; x=4

Câu này rất dễ theo đề bài x²  là x nhân x có nghĩa x nhân chính nó vậy ta có luôn x bằng 4 vì 4 nhân 4 trừ đi 4² bằng 0

Mik chỉ cần đáp án thôi nhé, ko cần vẽ trục số

\(\dfrac{2x-3}{5}-x+2\ge\dfrac{x}{3}\)

\(\Leftrightarrow3\left(2x-3\right)-15\left(x+2\right)\ge5x\)

\(\Leftrightarrow6x-9-15x+30\ge5x\)

\(\Leftrightarrow6x-15x-5x\ge9+30\)

\(\Leftrightarrow-14x\ge-21\)

\(\Leftrightarrow x\le\dfrac{21}{14}\le\dfrac{3}{2}\)

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lâu rồi cũng không nhớ cách làm :v

a: \(\frac{25}{14x^2y}=\frac{25\cdot3\cdot y^4}{14x^2y\cdot3y^4}=\frac{75y^4}{42x^2y^5}\)

\(\) \(\frac{14}{21xy^5}=\frac{14\cdot2\cdot x}{21xy^5\cdot2x}=\frac{28x}{42x^2y^5}\)

b: \(\frac{11}{102x^4y}=\frac{11\cdot y^2}{102x^4y\cdot y^2}=\frac{11y^2}{102x^4y^3}\)

\(\frac{3}{34xy^3}=\frac{3\cdot3\cdot x^3}{34xy^3\cdot3x^3}=\frac{9x^3}{102x^4y^3}\)

c: \(\frac{3x+1}{12xy^4}=\frac{\left(3x+1\right)\cdot3\cdot x}{12xy^4\cdot3x}=\frac{9x^2+3x}{36x^2y^4}\)

\(\frac{y-2}{9x^2y^3}=\frac{\left(y-2\right)\cdot4y}{4y\cdot9x^2y^3}=\frac{4y^2-8y}{36x^2y^4}\)

d: \(\frac{1}{6x^3y^2}=\frac{1\cdot6\cdot y^2}{6x^3y^2\cdot6y^2}=\frac{6y^2}{36x^3y^4}\)

\(\frac{x+1}{9x^2y^4}=\frac{\left(x+1\right)\cdot4\cdot x}{9x^2y^4\cdot4x}=\frac{4x^2+4x}{36x^3y^4}\)

\(\frac{x-1}{4xy^3}=\frac{\left(x-1\right)\cdot9\cdot x^2y}{4xy^3\cdot9x^2y}=\frac{9x^3y-9x^2y}{36x^3y^4}\)

e: \(\frac{3+2x}{10x^4y}=\frac{\left(2x+3\right)\cdot12\cdot y^4}{10x^4y\cdot12y^4}=\frac{24xy^4+36y^4}{120x^4y^5}\)

\(\frac{5}{8x^2y^2}=\frac{5\cdot15\cdot x^2y^3}{8x^2y^2\cdot15x^2y^3}=\frac{75x^2y^3}{120x^4y^5}\)

\(\frac{2}{3xy^5}=\frac{2\cdot40\cdot x^3}{3xy^5\cdot40x^3}=\frac{80x^3}{120x^4y^5}\)

f: \(\frac{4x-4}{2x\left(x+3\right)}=\frac{\left(4x-4\right)\cdot3\cdot\left(x+1\right)}{2x\left(x+3\right)\cdot3\cdot\left(x+1\right)}=\frac{12\left(x-1\right)\left(x+1\right)}{6x\left(x+3\right)\left(x+1\right)}=\frac{12x^2-12}{6x\left(x+3\right)\left(x+1\right)}\)

\(\frac{x-3}{3x\left(x+1\right)}=\frac{\left(x-3\right)\cdot2\cdot\left(x+3\right)}{3x\left(x+1\right)\cdot2\cdot\left(x+3\right)}=\frac{2\left(x^2-9\right)}{6x\left(x+1\right)\left(x+3\right)}\)

g: \(\frac{2x}{\left(x+2\right)^3}=\frac{2x\cdot2x}{2x\left(x+2\right)^3}=\frac{4x^2}{2x\left(x+2\right)^3}\)

\(\frac{x-2}{2x\left(x+2\right)^2}=\frac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}=\frac{x^2-4}{2x\left(x+2\right)^3}\)

h: \(\frac{5}{3x^3-12x}=\frac{5}{3x\left(x^2-4\right)}=\frac{5}{3x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{5\cdot2\cdot\left(x+3\right)}{3x\left(x-2\right)\left(x+2\right)\cdot2\cdot\left(x+3\right)}=\frac{10x+30}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)

\(\frac{3}{\left(2x+4\right)\left(x+3\right)}=\frac{3}{2\left(x+2\right)\left(x+3\right)}\)

\(=\frac{3\cdot3\cdot x\left(x-2\right)}{3x\left(x-2\right)\cdot2\left(x+2\right)\left(x+3\right)}=\frac{9x^2-18x}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x-y\right)}{\left(x-y\right)^2x\left(x+y\right)}=\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{x\left(x-y\right)^2\left(x+y\right)}=\dfrac{x^2+y^2}{x}\)

\(9-\left(x-y\right)^2\)

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

\(=\left(3-x+y\right)\left(3+x-y\right)\)

____

\(\left(x-y\right)^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

____

\(\left(x+2\right)^2-y^2\)

\(=\left[\left(x+2\right)-y\right]\left[\left(x+2\right)+y\right]\)

\(=\left(x-y+2\right)\left(x+y+2\right)\)

____

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

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

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

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

____

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

____
\(\left(2xy+1\right)^2-\left(2x+y\right)^2\)

\(=\left(2xy+1-2x-y\right)\left(2xy+1+2x+y\right)\)

\(=\left[2x\left(y-1\right)-\left(y-1\right)\right]\left[2x\left(y+1\right)+\left(y+1\right)\right]\)

\(=\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\left(y+1\right)\)

chia ra dùm

làm các bài đại trc nha

ĐK: \(x\ne0\)

\(\frac{20x^2-15x}{5x}+\left(\frac{12-9x}{3}\right)=15\)

\(\Leftrightarrow4x-3+4-3x=15\Leftrightarrow x=14\)

\(x^2-3x+xy-3x\)

\(=x^2+xy-6x\)

\(=x.\left(x+y-6\right)\)