đợi tí

a) 9(2x+2)=144

18x +18=144

18x = 126

x = 7

Vậy x = 7m

b) 6x+15 = 75

6x = 60

x = 10

Vậy x = 10m

c) 12x+24 = 168

12x = 144

x =12

Vậy x = 12m.

4

mình ấn lộn bạn nhé ra -2

a).

\(x^5+x+1=\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^3-x^2\right)\)

b).\(x^8+x^7+1=\left(x^8+x^7+x^6\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

d).

\(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

e).

\(x^8+x^4+1=x^8+2x^4+1-x^4\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)

c).

\(x^5-x^4-1=x^5-x^3-x^2-\left(x^4-x^2-x\right)+x^3-x-1\\ \left(x^3-x-1\right)\left(x^2-x+1\right)\)

c)(x²+x)²-2(x²+x)-15

đặt x²+x=a ta có

a²-2a-15

=a²+3a-5a-15

=(a²+3a)-(5a+15)

=a(a+3)-5(a+3)

=(a+3)(a-5)

thay a=x²+x

(x²+x+3)(x²+x-5)

a,(5x-2y)(x²-xy+1)=5x³-5x²+5x-2yx²+2xy²-2y

=5x³-7x²y+2xy²+5x-2y

b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x²-4.\(\left(\dfrac{1}{2}x-5\right)\)

=\(\dfrac{1}{2}x^3-5x^2-2x+20\)

c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)

=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)

=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)

=\(-5x+4x-15\)

=\(-x-15\)

Chúc bạn học tốt(mỏi tay quá)

Diện tích đáy lăng trụ là:

\(S=\dfrac{1}{2}\cdot2x=x\left(cm^2\right)\)

\(V=S\cdot h\)

=>x=V/h=3(cm)

\(x^5+y^5=\left(x+y\right)^3\)

\(pt\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=\left(x+y\right)^3\)

\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-\left(x+y\right)^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2\right)=0\)

*)Xét \(x+y=0\Rightarrow x=-y\)

\(\Rightarrow\left(-y\right)^5+y^5=\left(-y+y\right)^3\Rightarrow0=0\) (luôn đúng)

*)Xét \(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2=0\)

+)Xét \(x=0 \Rightarrow y=±1; y=0 \Rightarrow x=±1\)

Và \(x=y=0\)

+)Xét \(x;y\ne 0\) ta có:\(x^4-x^3y+x^2y^2-xy^3+y^4-x^2-2xy-y^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^4+y^4-x^3y-xy^3=0\\x^2y^2-x^2-2xy-y^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2\left(x^2+xy+y^2\right)=0\\\left(xy+x+y\right)\left(xy-x-y\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\\left(x^2+2x\right)\left(x^2-2x\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x^2\left(x-2\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=2;x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=y=-2\\x=y=2\end{matrix}\right.\)

Trình bày chắc sai chỗ nào đó nhưng kq thì đúng 100%-WF đã chứng minh :V