( 3x - 2⁴ ) × 7³ = 2×7⁴
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\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)
Answer:
\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)
\(\Rightarrow6x^2-6x^2-5x+6=7\)
\(\Rightarrow-5x+6=7\)
\(\Rightarrow-5x=1\)
\(\Rightarrow x=\frac{-1}{5}\)
\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)
\(\Rightarrow5x.19-240x^2+15x=-100\)
\(\Rightarrow95x-240x^2+15x=-100\)
\(\Rightarrow-240x^2+110x+100=0\)
\(\Rightarrow-24x^2-11x-10=0\)
\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)
\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)
\(\left(3x-5\right)\left(7-5x\right)-\left(5x-2\right)\left(2-3x\right)=4\)
\(\Rightarrow\left(21x-15x^2-35+25x\right)-\left(10x-15x^2-4+6x\right)-4=0\)
\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)
\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)
\(\Rightarrow30x-35=0\)
\(\Rightarrow x=\frac{7}{6}\)
a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)
=>-5x=1
hay x=-1/5
b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)
\(\Leftrightarrow-180x^2+50x+100=0\)
hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)
c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)
=>30x-31=4
=>30x=35
hay x=7/6
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
\(\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)
\(=\left(3x+2-2x+7\right)^2\)
\(=\left(x+9\right)^2\)