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a: =4x^2-4x+1+9
=(2x-1)^2+9>=9
Dấu = xảy ra khi x=1/2
b: =2(x^2+3x)
=2(x^2+3x+9/4-9/4)
=2(x+3/2)^2-9/2>=-9/2
Dấu = xảy ra khi x=-3/2
c: =x^2-x+1/4-1/4
=(x-1/2)^2-1/4>=-1/4
Dấu = xảy ra khi x=1/2
a. A=4x-x2+3= 7-(x2-4x+4)=7-(x-2)2
Nhận thấy -(x-2)2\(\le0\forall x\)
=> 7-(x-2)2\(\le7\forall x\)
Dấu "=" xảy ra khi x-2=0=>x=2
Vậy max A=7 <=>x=2
b. B= -x2+6x-11= -2-(x2-6x+9)=-2-(x-3)2
Nhận thấy -(x-3)2\(\le0\forall x\)
=> -2-(x-3)2 \(\le-2\forall x\)
Dấu "=" xảy ra khi x-3=0 => x=3
Vậy max B=-2 <=> x=3
Giải phương trình
a, 5x(x-4)-5x2 = 2 (11-x)
\(\Leftrightarrow5x^2-20x-5x^2=22-2x\)
\(\Leftrightarrow-18x=22\)
\(\Leftrightarrow x=\frac{-22}{18}\)
b, \(\frac{3}{x-3}-\frac{2}{x+3}=\frac{4x}{x^2-9}\left(x\ne\pm3\right)\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4x}{x^2-9}\)
\(\Rightarrow3x+9-2x+6=4x\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\left(tm\right)\)
Kl: a,.........
b,.........
a) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)
\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)
\(\Leftrightarrow x^2-11x+3x-33=14x-154\)
\(\Leftrightarrow x^2-22x+121=0\)
\(\Leftrightarrow\left(x-11\right)^2=0\Leftrightarrow x=11\)
Vậy .......
b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)
\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)
\(\Leftrightarrow2x^2=-15\)(vô lí)
Vậy pt vô nghiệm
c) \(\frac{3-2x}{x+1}=2+\frac{1-4x}{x-2}\)
\(\Leftrightarrow\left(3-2x\right)\left(x-2\right)=2\left(x+1\right)\left(x-2\right)+\left(1-4x\right)\left(x+1\right)\)
\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)
\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)
Vậy.........
(gửi trước 3 câu)
d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
Vậy x=-5/2
e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)
f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)
a: Ta có: \(A=x^2-7x+11\)
\(=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{5}{4}\)
\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{7}{2}\)
b: ta có: \(A=9x^2+6x+11\)
\(=9x^2+6x+1+10\)
\(=\left(3x+1\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{3}\)
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
Nhân chéo, chuyển vế đưa về dạng pt bậc 2, xét đenta cho nó >=0 rồi giải
\(A=\frac{4x-11}{x^2+3}=\frac{4x^2+4x+1-4x^2-12}{x^2+3}=\frac{\left(2x+1\right)^2-4\left(x^2+3\right)}{x^2+3}=\frac{\left(2x+1\right)^2}{x^2+3}-4\)
Phân số \(\frac{\left(2x+1\right)^2}{x^2+3}\ge0\forall x\Rightarrow A=\frac{\left(2x+1\right)^2}{x^2+3}-4\ge-4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(A_{min}=-4\Leftrightarrow x=-\frac{1}{2}\)
Chúc bạn học tốt.