\(\Rightarrow\left|3x+1\right|=4+\dfrac{1}{2}=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{9}{2}\\3x+1=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{2}\\3x=-\dfrac{11}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{11}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left|3x+1\right|=\dfrac{9}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=\dfrac{9}{2}\\3x+1=-\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{11}{6}\end{matrix}\right.\)

\(\frac{3}{8}\times x+\frac{1}{4}\times x=\frac{2}{5}\)

\(\Rightarrow x\times\left(\frac{3}{8}+\frac{1}{4}\right)=\frac{2}{5}\)

\(\Rightarrow x\times\frac{5}{8}=\frac{2}{5}\)

\(\Rightarrow x=\frac{16}{25}\)

~Std well~

#Thạc_Trân

bn ơi / là j bn

\(\frac{3}{8}\times x+\frac{1}{4}\times x=\frac{2}{5}\)

\(\left(\frac{3}{8}+\frac{1}{4}\right)\times x=\frac{2}{5}\)

\(\frac{5}{8}\times x=\frac{2}{5}\)

Đến đây tự làm :v

1/4 . x - (x - 4/5) - 1/2 = 2x - 6

1/4 x - x + 4/5 - 1/2 = 2x - 6

1/4 x - x - 2x = -6 - 4/5 + 1/2

-11/4 x = -63/10

x = -63/10 : (-11/4)

x = 126/55

x = 126/55

1.

Đặt \(x+y=a\Rightarrow y=a-x\)

\(\Rightarrow x^2+2x\left(a-x\right)-14\left(a-x\right)-10x+3\left(a-x\right)^2+27=0\)

\(\Leftrightarrow2x^2-4\left(a+1\right)x+3a^2-10a+27=0\)

\(\Delta'=4\left(a+1\right)^2-2\left(3a^2-10a+27\right)\ge0\)

\(\Leftrightarrow-a^2+14a-25\ge0\)

\(\Rightarrow7-2\sqrt{6}\le a\le7+2\sqrt{6}\)

\(\Rightarrow-10-2\sqrt{6}\le P\le-10+2\sqrt{6}\)

2. Chắc đề là \(a;b>0\) (đảm bảo mẫu dương) chứ ko phải \(a.b>4\)

\(M\ge\dfrac{\left(a+b\right)^2}{a+b-8}=\dfrac{\left(a+b-8+8\right)^2}{a+b-8}=\dfrac{\left(a+b-8\right)^2+16\left(a+b-8\right)+64}{a+b-8}\)

\(M\ge a+b-8+\dfrac{64}{a+b-8}+16\ge2\sqrt{\dfrac{64\left(a+b-8\right)}{a+b-8}}+16=32\)

Dấu "=" xảy ra khi \(a=b=8\)

\(a)\) \(2x-5=21\)

\(\Leftrightarrow\) \(2x=21+5\)

\(\Leftrightarrow\) \(2x=26\)

\(\Leftrightarrow\) \(x=26:2\)

\(\Leftrightarrow\) \(=13\)

\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)

\(\Leftrightarrow\) \(x=\frac{1}{3}\)

\(c)\) \(\frac{7}{8}-\frac{1}{2}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{7}{8}-\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{1}{24}\)

\(\Leftrightarrow\) \(x=\frac{1}{12}\)

1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)

2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)

3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)

4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)