6x(15+5x)=300
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\(\Leftrightarrow6x+4-5\left|x\right|=16\\ \Leftrightarrow5\left|x\right|=6x-12\\ \Leftrightarrow\left[{}\begin{matrix}5x=6x-12\left(x\ge0\right)\\5x=12-6x\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(tm\right)\\x=\dfrac{12}{11}\left(ktm\right)\end{matrix}\right.\)
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
Ta có:
\(2x^3-5x^2+6x-15\)
\(=\left(2x^3-5x^2\right)+\left(6x-15\right)\)
\(=x^2\left(2x-5\right)+3\left(2x-5\right)\)
\(=\left(x^2+3\right)\left(2x-5\right)\)
\(\Rightarrow\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)=x^2+3\)
+) ta có : \(E=3x^2-6x+15=3\left(x^2-2x+1\right)+12\)
\(=3\left(x-1\right)^2+12\ge12\) \(\Rightarrow E_{min}=12\) khi \(x=1\)
+) ta có : \(F=5x^2+6x-12=5\left(x^2+\dfrac{6}{5}x+\dfrac{9}{25}\right)-\dfrac{69}{5}\)
\(=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{69}{5}\ge\dfrac{-69}{5}\) \(\Rightarrow F_{min}=-\dfrac{69}{5}\) khi \(x=\dfrac{-3}{5}\)
+) ta có : \(G=4x^2-4x+25=4\left(x^2-x+\dfrac{1}{4}\right)+24\)
\(=4\left(x-\dfrac{1}{2}\right)^2+24\ge24\) \(\Rightarrow G_{min}=24\) khi \(x=\dfrac{1}{2}\)
+) ta có : \(H=9x^2+6x^2+4=15x^2+4\ge4\)
\(\Rightarrow H_{min}=4\) khi \(x=0\)
Tìm GTNN
E=3x^2-6x+15
F= 5x^2+6x-12
G=4x^2-4x+25
H=9x^2+6x^2+4
x = 7
6 × (15 + 5x) = 300
15 + 5x = 300 : 6
15 + 5x = 50
5x = 50 - 15
5x = 35
x = 35 : 5
x = 7