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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.
giúp mk vs nha
\(3f\left(x\right)+2f\left(1-x\right)=2x+9\)
\(\left\{\begin{matrix}3f\left(2\right)+2f\left(-1\right)=2.2+9=13\left(1\right)\\3f\left(-1\right)+2f\left(2\right)=2.\left(-1\right)+9=7\left(2\right)\end{matrix}\right.\)
Lấy (1) nhân 3 trừ đi (2) nhân 2:
\(\left(3.3-2.2\right)f\left(2\right)+\left(6-6\right)f\left(-1\right)=13.3-7.2\)
\(f\left(2\right)=\frac{39-14}{9-4}=\frac{25}{5}=5\)
Câu hỏi của Phạm Mai Chi - Toán lớp 8 - Học toán với OnlineMath
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 )
Để A \(⋮\) B thì \(x^n\ge x^3\) \(\Rightarrow n\ge3\)
Để M \(⋮\) N thì \(y^n\ge y^2\Rightarrow n\ge2\)
a, A= 5\(x^ny^3\)
B= 4\(x^3y\)
=> A\(⋮\)B -> n \(\ge\)3
b, làm tương tự như trên
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).
\(x^8+x^4+1=x^8+2x^4+1-x^4\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
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)
Ta có:
A N A B = 3 9 = 1 3 , A M A C = 6 18 = 1 3 ⇒ = A M A C = 1 3
Xét ΔANM và ΔABC có:
A N A B = A M A C (chứng minh trên)
A chung
⇒ ΔANM ~ ΔABC (c - g - c)
⇒ A N A B = A M A C = M N C B = 1 3 ⇒ x 15 = 1 3 ⇒ x = 15 3 = 5
Đáp án: B