\(a.\left(x^2-2\right)^2=\left(x^2\right)^2-2\cdot x^2\cdot2+2^2=x^4-4x^2+4\\ b.\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\\ c.\left(\dfrac{1}{2}-x\right)^2=\left(\dfrac{1}{2}\right)^2-2\cdot x\cdot\dfrac{1}{2}+x^2=\dfrac{1}{4}-x+x^2\\ d.\left(2a-1\right)^2=\left(2a\right)^2-2\cdot2a\cdot1+1^2=4a^2-4a+1\\ e.\left(\dfrac{1}{3}a+2\right)^2=\left(\dfrac{1}{3}a\right)^2+2\cdot\dfrac{1}{3}a\cdot2+2^2=\dfrac{1}{9}a^2+\dfrac{4}{3}a+4\\ f.\left(2a-\dfrac{2}{3}\right)^2=\left(2a\right)^2-2\cdot2a\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2=4a^2-\dfrac{8}{3}a+\dfrac{4}{9}\)

Xét ΔABC vuông tại A có \(\widehat{ABC}+\widehat{ACB}=90^0\)

=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)

=>\(\widehat{OBC}+\widehat{OCB}=45^0\)

Xét ΔOBC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)

=>\(\widehat{BOC}+45^0=180^0\)

=>\(\widehat{BOC}=135^0\)

\(x^{10}=1^x\\ =>x^{10}=1\\ =>x^{10}=\left(\pm1\right)^{10}\\ =>x=\pm1\)

Vậy: ...

ΔABC cân tại A

=>\(\widehat{BAC}=180^0-2\cdot\widehat{ABC}=100^0\)

AD là phân giác góc ngoài tại đỉnh A

=>\(\widehat{CAD}=\dfrac{180^0-\widehat{BAC}}{2}=40^0\)

=>\(\widehat{CAD}=\widehat{ACB}\left(=40^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên AD//BC

Đặt: \(\dfrac{1}{2x-y}=a;\dfrac{1}{x+y}=b\left(2x\ne y;x\ne-y\right)\)

Hpt trở thành:

\(\left\{{}\begin{matrix}3a-6b=1\\a-b=0\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}3a-6a=1\\a=b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-3a=1\\a=b\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{3}\\b=a=-\dfrac{1}{3}\end{matrix}\right.\) 

\(=>\left\{{}\begin{matrix}\dfrac{1}{2x-y}=\dfrac{1}{-3}\\\dfrac{1}{x+y}=\dfrac{1}{-3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=-3\\x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-6\\x+y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-6}{3}=-2\\-2+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3+2=-1\end{matrix}\right.\)

Xét 2 ΔABO và ΔADO ta có:

\(\widehat{BAO}=\widehat{DAO}\) (AD là phân giác của góc BAC) 

\(OA\) chung

\(\widehat{AOB}=\widehat{AOD}\left(gt\right)\) 

\(=>\Delta ABO=\Delta ADO\left(g.c.g\right)\) 

\(=>\widehat{B}=\widehat{D_1}\) (hai góc tương ứng) 

a) 575-(6x+70)=455

6x+70=575-455

6x+70=120

6x=120-70

6x=50

x=50:6

x=\(\dfrac{25}{3}\) \(\notinℕ\)

Vậy không có giá trị tự nhiên x thỏa mãn đề

b) 315+(125-x)=435

125-x=435-315

125-x=120

x=125-120

x=5 (nhận)

c) 3x+28=88

3x=88-28

3x=60

x=60:3

x=20

\(a.575-\left(6\cdot x+70\right)=455\\ =>6\cdot x+70=575-455\\ =>6\cdot x+70=120\\ =>6\cdot x=120-70\\ =>6\cdot x=50\\ =>x=\dfrac{50}{6}=\dfrac{25}{3}\left(L\right)\\ b.315+\left(125-x\right)=435\\ =>125-x=435-315=125\\ =>x=125-125=0\\ c.3\cdot x+28=88\\ =>3\cdot x=88-28\\ =>3\cdot x=60\\ =>x=\dfrac{60}{3}=20\\ x-105:21=15\\ =>x-5=15\\ =>x=15+5=20\\ e.\left(x-105\right):21=15\\ =>x-105=21\cdot15\\ =>x-105=315\\ =>x=315+105\\ =>x=420\)

\(x^2+2x+10=\left(x^2+2x+1\right)+9\\ =\left(x+1\right)^2+9\ge9>0\forall x\inℝ\left(Vì:\left(x+1\right)^2\ge0\forall x\inℝ\right)\)

Bạn xem lại đề nhé.

Đặt: \(3x^2-5x-7=0\) 

\(\Delta=\left(-5\right)^2-4\cdot3\cdot\left(-7\right)=109>0\)

\(x_1=\dfrac{-\left(-5\right)+\sqrt{109}}{2\cdot3}=\dfrac{5+\sqrt{109}}{6}\)

\(x_2=\dfrac{-\left(-5\right)-\sqrt{109}}{2\cdot3}=\dfrac{5-\sqrt{109}}{6}\) 

=> \(3x^2-5x-7=\left(x-\dfrac{5+\sqrt{109}}{6}\right)\left(x-\dfrac{5-\sqrt{109}}{6}\right)\)

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

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

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

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

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