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b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{x^2+5x+8}{5}\)
\(=\dfrac{\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{7}{4}}{5}\)
\(=\dfrac{\left(x+\dfrac{5}{2}\right)^2}{5}+\dfrac{7}{20}\)
Vì \(\dfrac{\left(x+\dfrac{5}{2}\right)^2}{5}\ge0,\text{∀x}\)
⇒ \(A\ge\dfrac{7}{20},\text{∀x}\)
Min \(A=\dfrac{7}{20}\)⇔\(x=-\dfrac{5}{2}\)
\(A=\dfrac{x^2+5x+8}{5}=\dfrac{\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{7}{4}}{5}=\dfrac{\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}}{5}\ge\dfrac{\dfrac{7}{4}}{5}=\dfrac{7}{4}.\dfrac{1}{5}=\dfrac{7}{20}\)-GTNN của A là \(\dfrac{7}{20}\Leftrightarrow x+\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{-5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
⇔ (x - 1)(5x + 3) - (3x - 8)(x - 1) = 0
⇔ (x - 1)(5x + 3 - 3x + 8) = 0
⇔ (x - 1)(2x + 11) = 0
⇔\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-11}{2}\end{matrix}\right.\)
Vậy S = {1; \(\frac{-11}{2}\)}
b) 3x(25x + 15) - 35(5x + 3) = 0
⇔ 15x(5x + 3) - 35(5x + 3) = 0
⇔ 5(3x - 7)(5x + 3) = 0
⇔ \(\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{-3}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{7}{3};\frac{-3}{5}\)}
a/ \(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{11}{2}\end{matrix}\right.\)
b/ \(3x.5\left(5x+3\right)-5.7\left(5x+3\right)=0\)
\(\Leftrightarrow5\left(3x-7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{3}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}15x-35=0\\5x+3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-3}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{-3}{5}\right\}\)
3x(25x + 15) - 35(5x + 3) = 0
<=> 15x(5x + 3) - 35(5x + 3) = 0
<=> (5x + 3)(15x - 35) = 0
<=> 5(5x + 3)(3x - 7) = 0
<=> 5x + 3 = 0 hay 3x - 7 = 0 (vì 5 \(\ne\)0)
<=> 5x = -3 I <=> 3x = 7
<=> x =\(\frac{-3}{5}\)I <=> x = \(\frac{7}{3}\)
Vậy S = {\(\frac{-3}{5}\); \(\frac{7}{3}\)}
\(A=5x^2-25x+35+7y^8\)
\(=5\left(x^2-5x+7\right)+7y^8\)
\(=5\left(x^2-5x+\frac{25}{4}+\frac{3}{4}\right)+7y^8\)
\(=5\left[\left(x-\frac{5}{2}\right)^2+\frac{3}{4}\right]+7y^8\)
\(=5\left(x-\frac{5}{2}\right)^2+\frac{15}{4}+7y^8\ge\frac{15}{4}\)
\(\Leftrightarrow x=\frac{5}{2};y=0\)