Ôn tập toán 8

\(2x^2+3x-2=0\)

\(\Rightarrow2x^2-x+4x-2=0\)

\(\Rightarrow x.\left(2x-1\right)+2.\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right).\left(x+2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+2=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=1\\x=-2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-2\end{array}\right.\)

\(2x^2+3x-2=0\)

\(\Leftrightarrow2x^2+4x-x-2=0\)

\(\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-2\end{array}\right.\)

\(2x^2+3x-2=0\)

\(\Delta=3^2-4.2.\left(-2\right)=25\)

\(X_1=\frac{-3+\sqrt{25}}{4}=\frac{1}{2}\)

\(X_2=\frac{-3-\sqrt{25}}{4}=-2\)

\(P=\left(x^2+2.x.2+2^2\right)+1\)

\(\Rightarrow P=\left(x+2\right)^2+1\)

TA có

(x+2)^2 lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left(x+2\right)^2+1\ge1\)

Dấu " = " xảy ra khi x= - 2

Vậy MIN_P=1 khi x = - 2

b)

\(Q=-\left(x^2-4x+2^2\right)+9\)

\(\Rightarrow Q=-\left(x-2\right)^2+9\)

TA có

(x - 2)^2 lớn hơn hoặc bằng 0 với mọi x

=> - (x - 2)^2 bé hơn hoặc bằng 0 

\(\Rightarrow-\left(x-2\right)^2+9\le9\)

Dấu " = " xảy ra khi x=2

Vậy MAX_Q=9 khi x = 2

a) \(P=x^2+4x+5=x^2+4x+4+1=\left(x+2\right)^2+1\) 

Vì \(\left(x+2\right)^2\ge0\) với mọi x

=> \(\left(x+2\right)^2+1\ge1\) với mọi x

=> Pmin = 1

Dấu "=" xảy ra <=> x+2 = 0 <=> x = -2

b) \(Q=-x^2+4x+5=-\left(x^2-4x+4\right)+9\)

  \(=-\left(x-2\right)^2+9\)

Vì \(-\left(x-2\right)^2\le0\) với mọi x

=> \(-\left(x-2\right)^2+9\le9\)

=> Qmax = 9

Dấu "=" xảy ra <=> x - 2 = 0  <=> x = 2

\(=\left(x-y\right)^2-2\left(x-y\right)z^3+\left(z^3\right)^2\)

\(=\left(z-y-z\right)^2\)

\(=\left(x-y\right)^2-2\left(x-y\right).z^3+\left(z^3\right)^2\)

\(=\left(x-y-z\right)^2\)

bài có nghĩa là sao ??? Phân tích hay tính z bn ???

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\left(\frac{1-x}{2002}+1\right)+\left(\frac{-x}{2003}+1\right)\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow\) \(x=2003\) 

↔ \(\frac{2-x}{2001}+1\)\(=\left(\frac{1-x}{2002}+1\right)+\left(\frac{x}{2003}+1\right)\)

↔ \(\frac{2003-x}{2001}\) \(=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

↔ \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

↔ x = 2003

\(\frac{x-1}{2}+\frac{x-1}{4}=1-\frac{2\left(x-1\right)}{3}\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{2}{3}\right)=1\)

\(\Leftrightarrow\left(x-1\right)\frac{17}{12}=1\)

\(\Leftrightarrow x-1=\frac{12}{17}\)

\(\Leftrightarrow x=\frac{29}{27}\)

Lần này thì đúng rồi :

\(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}-\frac{1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}+\frac{x}{3}=4+\frac{1}{6}\)

\(\Leftrightarrow\frac{4x}{3}=\frac{25}{6}\)

\(\Leftrightarrow x=\frac{25}{8}\)

\(=\frac{4x}{6}+\frac{2x-1}{6}\)

\(=\frac{6x-1}{6}\)

\(=\frac{6x}{6}-\frac{1}{6}=x-\frac{1}{6}\)

Đề sai

Cho mình xin lỗi đề là gì nè : \(\frac{2x}{3}+\frac{2x-1}{6}\)