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

\(\Rightarrow3x^2-3x+2x-2-3\left(x^2-2x+x-2\right)=4\)

\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)

\(\Rightarrow2x+4=4\)

\(\Rightarrow2x=0\Leftrightarrow x=0\)

\(\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)

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

\(=x^2-2x-4x+8-x^2+3x+x-3\)

\(=-2x+5\)

(x - 4).(x - 2) - (x - 1).(x - 3)

= x² - 2x - 4x + 8 - ( x² - 3x - x + 3 )

= x² - 2x - 4x + 8 - x² + 3x + x - 3

= 5 - 2x 

......

\(pt\Leftrightarrow\hept{\begin{cases}\frac{1}{2}xy+\frac{3}{2}x+y+3=\frac{1}{2}xy+50\\\frac{1}{2}xy-x-y+2=\frac{1}{2}xy-32\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{3}{2}x+y=47\\-x-y=-34\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=26\\y=8\end{cases}}\)

Vậy pt có một nghiệm duy nhất (x;y) = (26;8).

\(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}-\frac{x^2-y^2}{xy}-\frac{y^2}{xy+y^2}\right)\)\(\left(\frac{x+y}{x^2+xy+y^2}\right)\)

ĐK: \(\hept{\begin{cases}x,y\ne0\\x\ne-y\end{cases}}\)

\(A=\frac{2}{x}-\frac{x^2y-\left(x-y\right)\left(x+y\right)^2-xy^2}{xy\left(x+y\right)}.\frac{x+y}{x^2+xy+y^2}\)

\(A=\frac{2}{x}+\frac{x^3-y^3}{xy\left(x+y\right)}.\frac{x+y}{x^2+xy+y^2}\)

\(A=\frac{2}{x}+\frac{x-y}{xy}\)

\(A=\frac{2y+x-y}{xy}\)

\(A=\frac{x+y}{xy}\)

1) \(\widehat{A}+\widehat{D}=180^O\)

=> \(\widehat{A}=180^O-60^O=120^O\)

2) \(\frac{\widehat{B}}{\widehat{D}}=\frac{4}{5}\)=> \(\widehat{B}=60.\frac{4}{5}=48^O\)

Ta có: \(\widehat{B}+\widehat{C}=180^o\)

        => \(\widehat{C}=180^o-48^{^{ }o}=132^o\)

không biết mik giải đúng ko mà đáp án nó không đúng thực tế lắm

Gọi giao điểm các đường phân giác trong tứ giác ABCD lần lượt là M, N, P, Q như hình vẽ bên trên.

Xét tam giác APB có: \(\widehat{APB}=180^o-\widehat{PAB}-\widehat{PBA}=\frac{360^o-\widehat{DAB}-\widehat{CBA}}{2}\)

Tương tự xét tam giác MCD ta cũng có:

\(\widehat{DMC}=\frac{360^o-\widehat{ADC}-\widehat{BCD}}{2}\)

Suy ra \(\widehat{QMN}+\widehat{QPN}=\frac{360^o-\widehat{ADC}-\widehat{BCD}}{2}+\frac{360^o-\widehat{DAB}-\widehat{ABC}}{2}\)

\(=\frac{720^o-360^o}{2}=180^o\)

Do tổng 4 góc trong một tứ giác bằng 360^o nên ta cũng có \(\widehat{MQP}+\widehat{MNP}=360^o-180^o=180^o\)

Vậy tứ giác MNPQ có các góc đối bù nhau.

a, x = 79 => x + 1 = 80

Ta có:\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)

\(=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)

\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+x^2+x+15\)

\(=x+15=79+15=94\)

Còn lại tương tự

\(Q_{\left(x\right)}=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

\(x^4+81\)

\(=x^4+3^4\)

\(=\left(x^2+3^2\right)^2-2x^23^2\)

\(=\left(x^2+\sqrt{2}x3+3^2\right)\left(x^2-\sqrt{2}x3+3^2\right)\)

nguồn gg

\(x^4+81\)

\(=x^4+18x^2+81-18x^2\)

\(=\left(x^2+9\right)^2-18x^2\)

\(=\left(x^2-3\sqrt{2}x+9\right)\left(x^2+3\sqrt{2}x+9\right)\)