\(a\left(a-b\right)\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a+b\right)\left[a\left(a-b\right)-\left(a^2-ab+b^2\right)\right]\)

\(=\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)\)

\(=\left(a+b\right)\left(-b^2\right)\)

\(92.4-27=\left(x+350\right):x+315\)

\(\Leftrightarrow368-27=\left(x+350\right):x+315\)

\(\Leftrightarrow341=\left(x+350\right):x+315\)

\(\Leftrightarrow\left(x+350\right):x=341-315\)

\(\Leftrightarrow\left(x+350\right):x=26\)

\(\Leftrightarrow\left(x:x\right)+\left(350:x\right)=26\)

\(\Leftrightarrow1+350:x=26\)

\(\Leftrightarrow350:x=26-1\)

\(\Leftrightarrow350:x=25\)

\(\Leftrightarrow x=\dfrac{350}{25}=14\)

Vậy x=14

a) Theo đề bài ta có:

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{x}{4}=\dfrac{z}{3}\) và \(x-y+z=50\)

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{x}{4}=\dfrac{z}{3}\Rightarrow\dfrac{x}{4}=\dfrac{y}{6};\dfrac{x}{4}=\dfrac{z}{3}\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{x-y+z}{4-6+3}=\dfrac{50}{1}=50\)

\(\dfrac{x}{4}=50\Rightarrow x=50.4=200\)

\(\dfrac{y}{6}=50\Rightarrow y=50.6=300\)

\(\dfrac{z}{3}=50\Rightarrow z=50.3=150\)

Vậy \(x=200,y=300,z=150\)

Gọi a là số học sinh của lớp 6A và \(a\in N\)

\(a⋮3\Rightarrow a\in B\left(3\right)\)

\(a⋮4\Rightarrow a\in B\left(4\right)\)

\(a⋮6\Rightarrow a\in B\left(6\right)\)

=> a là \(BCNN\left(3;4;6\right)\)

\(3=3\)

\(4=2^2\)

\(6=2.3\)

\(\Rightarrow BCNN\left(3;4;6\right)=2^2.3=12\)

\(\Rightarrow BC\left(12\right)=\left\{12;24;36;48;60;.....\right\}\)(I)

Ta có: \(30< a< 40\) (II)

Từ (I) và (II) \(\Rightarrow a=36\)

Vậy số học sinh của lớp 6A là 36 học sinh

a) \(4x^3-14x^2\)

\(=2x^2\left(2x-7\right)\)

b) \(5y^{10}+15y^6\)

\(=5y^6\left(y^4+3\right)\)

c) \(9x^2+15x^2y-21xy\)

\(=3x\left(3x+5xy-7y\right)\)

d) \(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(x-y\right)\left(5x+4y\right)\)

Tứ giác ABCD có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\text{ ( Tổng các góc tứ giác ) }\)

\(\Rightarrow\widehat{D}=360^o-\left(\widehat{A}+\widehat{B}+\widehat{C}\right)\)

\(\Rightarrow\widehat{D}=360^o-\left(90^o+40^o+70^o\right)\)

\(\Rightarrow\widehat{D}=360^o-200^o\)

\(\Rightarrow\widehat{D}=160^o\)

Vậy \(\widehat{D}=160^o\)

a)

b) E là trung điểm của BC nên \(CE=EB\)

Xét \(\Delta AEC\) và \(\Delta AEB\) ta có:

\(AC=AB=3cm\left(gt\right)\)

\(EC=EB\left(cmt\right)\)

\(AE:\text{ cạnh chung}\)

\(\Rightarrow\Delta AEC=\Delta AEB\left(c.c.c\right)\)

\(\Rightarrow\widehat{CAE}=\widehat{BAE}\text{ ( 2 góc tương ứng )}\)

\(\Rightarrow AE\text{ là tia phân giác của }\widehat{CAB}\left(dpcm\right)\)

Vì bốn điểm A, B, C, D thuộc đường tròn \(\left(O\right)\), nên \(OA=OB=OC=OD\)

Xét \(\Delta OAB\) và \(\Delta ODC\) có:

\(OA=OD\left(cmt\right)\)

\(OB=OC\left(cmt\right)\)

\(AB=DC\left(gt\right)\)

\(\Rightarrow\Delta OAB=\Delta ODC\left(c.c.c\right)\)

\(\Rightarrow\widehat{AOB}=\widehat{DOC}\text{ ( 2 góc tương ứng )}\)

\(\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x=3 hoặc x=1

Ta có:

\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{-8}{x\left(x-2\right)}\)

MTC: \(x\left(x-2\right)\left(x+2\right)\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{-8}{x\left(x-2\right)}=\dfrac{-8\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-8x-16}{x\left(x-2\right)\left(x+2\right)}\)