\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

\(=3x^2-3x-x+1\)

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

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

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

Vi tu giac ABCD co ^A = ^C = 90^o => ^B + ^D = 180^o

Kẻ phân giác DF , BE

Xét \(\Delta BEC\)vuông tại C nên \(\widehat{CBE}+\widehat{CEB}=90^o\)

\(\Rightarrow2\left(\widehat{CBE}+\widehat{CEB}\right)=180^o\)

\(\Rightarrow\widehat{CBA}+2\widehat{CEB}=180^o\)

Tuong tu \(\widehat{CDA}+2\widehat{AFD}=180^o\)

\(\Rightarrow\left(\widehat{CBA}+\widehat{CDA}\right)+2\left(\widehat{CEB}+\widehat{AFD}\right)=360^o\)

\(\Leftrightarrow180^o+2\left(\widehat{CEB}+\widehat{AFD}\right)=360^o\)

\(\Leftrightarrow\widehat{CEB}+\widehat{AFD}=90^o\)

\(\Rightarrow\widehat{CBE}=\widehat{AFD}\)(Cùng phụ \(\widehat{CEB}\))

\(\Rightarrow\widehat{ABE}=\widehat{AFD}\)(Phan giac)

\(\Rightarrow FD//\left(h\right)\equiv BE\left(dpcm\right)\)

Cảm ơn bạn Dương đã giúp mình làm nha!

\(a,x^2+2x-3\)

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

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

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

\(b,x^2-3x-10\)

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

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

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

\(c,x^2-13x+36\)

\(=x^2-4x-9x+36\)

\(=x\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-9\right)\left(x-4\right)\)

\(d,x^2+3x-18\)

\(=x^2+6x-3x-18\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x-3\right)\left(x+6\right)\)

Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)

Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Hay \(A=3\cdot2x\cdot2y\cdot2z\)

\(A=24xyz\)

\(VT=\left(x+y+z\right)^3-x^3-y^3-z^3\)

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

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

\(=\left(y+z\right)\left(y^2+2yz+z^2+3x^2+3xy+3xz-y^2+yz-z^2\right)\)

\(=\left(y+z\right)\left(3yz+3x^2+3xy+3xz\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)=VP\left(\text{ĐPCM}\right)\)

\(\frac{x-1}{2}\cdot\frac{x+1}{2}\cdot(4x-1)\)

\(=\frac{\left(x-1\right)\left(x+1\right)\left(4x-1\right)}{2\cdot2}\)

\(=\frac{(x^2-1)\left(4x-1\right)}{4}\)

\(=\frac{4x^3-x^2-4x+1}{4}\)

Ta có :

\(x^3+x^2z+y^2z-xyz+y^3\)

\(=x^3+y^3+x^2z+y^2z-xyz\)

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

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

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

\(=0\left(ĐPCM\right)\)