1, Tìm GTNN
A=x2+3x-1
B=2x2+5x+2
tìm gtnn (gtln) của
a) 4x2+12x+1 b) 4x2-3x+10
c)2x2+5x+10 d) x-x2+2
e) 2x-2x2 f) 4x2+2y2+4xy+4y+5
a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)
\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)
c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)
d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)
\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a: Ta có: \(4x^2+12x+1\)
\(=4x^2+12x+9-8\)
\(=\left(2x+3\right)^2-8\ge-8\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
b: Ta có: \(4x^2-3x+10\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)
c: Ta có: \(2x^2+5x+10\)
\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)
\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)
Tìm GTNN hoặc GTLN của các biểu thức sau:
a) A = x2 + 3x + 4
b) B = 2x2 - x + 1
c) C = 5x - x2 + 4
d) D = x2 + 5y2 - 2xy + 4y + 3
a: Ta có: \(A=x^2+3x+4\)
\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
GTNN:
A= x2+2x+5
B= x2_x+1
C= 5x2+5x+1
D= 3x2+4x+2
E= 1/2x2+x_1
F= 1/9x2+3x+2
\(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4\)
Kl: MinA = 4
\(B=x^2-x+1=\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
kl:.......
\(C=5x^2+5x+1=5\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+1-\dfrac{5}{4}=5\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
kl:.......
\(D=3x^2+4x+2=3\left(x^2+2\cdot\dfrac{2}{3}x+\dfrac{4}{9}\right)+2-\dfrac{4}{3}=3\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)
kl:......
\(E=\dfrac{1}{2}\cdot x^2+x-1=\dfrac{1}{2}\left(x^2+2x+1\right)-1-\dfrac{1}{2}=\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)
kl:............
\(F=\dfrac{1}{9}x^2+3x+2=\dfrac{1}{3}\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+2-\dfrac{1}{12}=\dfrac{1}{3}\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\)
kl:..........
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
bài 1 giải các bất phương trình sau
a, -x2 +5x-6 ≥ 0
b, x2-12x +36≤0
c, -2x2 +4x-2≤0
d, x2 -2|x-3| +3x ≥ 0
e, x-|x+3| -10 ≤0
bài 2 xét dấu các biểu thức sau
a,<-x2+x-1> <6x2 -5x+1>
b, x2-x-2/ -x2+3x+4
c, x2-5x +2
d, x-< x2-x+6 /-x2 +3x+4 >
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
a) (x2 - 2x)2 - 6x2 +12x + 9 = 0
b) (x2 + x + 1)(x2 + x + 2) = 12
c) (2x2 - 3x + 1)(2x2 + 5x + 1) - 9x2 = 0
a) Ta có: \(\left(x^2-2x\right)^2-6x^2+12x+9=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)(Vì \(x^2+x+5>0\forall x\))
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: S={-2;1}
2 ý a và b anh CTV nãy đã làm rồi nha, còn câu c này thì làm dài dòng+không chắc :VVV
c)\(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)\left(2x^2-3x+1+8x\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)^2+8x\left(2x^2-3x+1\right)+16x^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1+4x\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x^2-4x+1\right)=0\\\left(2x^2+6x+1\right)=0\end{matrix}\right.\)
Rồi đến đây tự giải nhé, không phân tích được thì bấm máy tính là ra nha:vv
Tất cả những bài này bạn đều có thể đặt ẩn phụ. Sau đó phân tích thành nhân tử để tìm nghiệm.
a) Đặt $x^2-2x=a$
b) Đặt $x^2+x+1=a$
c) Đặt $2x^2-3x+1=a$
Bài 1: Rút gọn rồi tính giá trị biểu thức:
a) A = 4x2.(-3x2 + 1) + 6x2.( 2x2 – 1) + x2 khi x = -1
b) B = x2.(-2y3 – 2y2 + 1) – 2y2.(x2y + x2) khi x = 0,5 và y = -1/2
Bài 2: Tìm x, biết:
a) 2(5x - 8) – 3(4x – 5) = 4(3x – 4) +11
b) 2x(6x – 2x2) + 3x2(x – 4) = 8
c) (2x)2(4x – 2) – (x3 – 8x2) = 15
Bài 3: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của biến x:
P = x(2x + 1) – x2(x+2) + x3 – x +3
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
\(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\\ P=2x^2+x-x^3-2x^2+x^3-x+3\\ P=3\left(đfcm\right)\)
a) (x2 - 5x)2 + 10(x2 - 5x) + 24 = 0
b) (2x + 1)2 - 2x - 1 = 2
c) x(x - 1)(x2 - x + 1) - 6 = 0
d) (x2 + 1)2 + 3x(x2 + 1) + 2x2 = 0
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
Cho biểu thức P= 1+ 3/x2+5x+6 : ( 8x2/ 4x3-8x2 - 3x/ 3x2-12 -1/x+2)
A) Rút gọn P
B) Tìm các giá trị của x để P= 0; P= 1
C) Tìm cã giá trị của x để P> 0
Cho biểu thức
Q= (2x-x2/ 2x2 +8 - 2x2/ 3x3-2x2+4x-8) (2/x2 + 1-x/x)
A) Rút gọn Q
B) Tìm giá trị nguyên của x để Q có giá trị nguyên
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
Làm tính nhân :
a) 2x. (x2 – 7x -3)
b) ( -2x3 + y2 -7xy). 4xy2
c)(-5x3).(2x2+3x-5)
d) (2x2 - xy+ y2).(-3x3)
e)(x2 -2x+3). (x-4) f) ( 2x3 -3x -1). (5x+2)
a: \(=2x^3-14x^2-6x\)
c: \(=-10x^5-15x^4+25x^3\)
a) 2x. (x2 – 7x -3)
= 2x3- 14x2- 6x
b) ( -2x3 + y2 -7xy). 4xy2
= -8x4y2+ 4xy4- 28x2y3
c)(-5x3).(2x2+3x-5)
= -10x5-15x4+25x3
d) (2x2 - xy+ y2).(-3x3)
=-6x5+ 3x4y -3x3y2
e)(x2 -2x+3). (x-4)
=x3-2x2+3x -4x2+8x-12
=x3-6x2+11x-12
f) ( 2x3 -3x -1). (5x+2)
=10x4-15x2-5x +4x3-6x-2
=10x4+4x3-15x2-11x-2