Dự đoán x = 2/5; y =4/7, giúp ta có được lời giải:D

\(A=\frac{5x}{2}+\frac{2}{5x}+\frac{7y}{2}+\frac{8}{7y}+\frac{1}{2}\left(x+y\right)\)

Đến đây đánh giá cô si + kết hợp giả thiết là xong:D

k ko biết

treen toán ko dc đưa những hình ảnh này. OK

khiếp thật

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 \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

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\(2x+11=0\Rightarrow x=\frac{-11}{2}\)

\(=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)= \(\left|2-5x\right|+\left|5x\right|\ge2+5x-5x=2\)

min A=2 \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow0\le x\le\frac{2}{5}}\)

Chuẩn đấy

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=−\(\frac{3}{5}\)

\(A=\dfrac{x^2+5x+8}{5}\)

\(=\dfrac{\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{7}{4}}{5}\)

\(=\dfrac{\left(x+\dfrac{5}{2}\right)^2}{5}+\dfrac{7}{20}\)

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 \(\dfrac{7}{20}\Leftrightarrow x+\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{-5}{2}\)

a)  x + 30 % x = − 1 , 3

x 1 + 3 10 = − 13 10 13 10 x = − 13 10 x = − 1

b)  1 3 x + 2 5 x − 1 = 0

1 3 x + 2 5 x − 2 5 = 0 11 15 x = 2 5 x = 2 5 : 11 15 x = 6 11

c)  3 x − 1 2 − 5 x + 3 5 = − x + 1 5

3 x − 3 2 − 5 x − 3 = − x + 1 5 x = − 3 2 − 3 − 1 5 x = − 47 10