1) Ham so nao sau day la ham so chan: A. y = \(\frac{x^2+\left|x\right|}{x+3}\) B. y= x3 + x C. y = x2 - 3x - 1 D. y = \(\frac{x^2}{2}+\frac{3}{x^2}\)
Những câu hỏi liên quan
a,ve do thi cac ham so y=\(\frac{-3}{2}\)x va y=\(\frac{2}{3}\)x tren cung 1 he truc toa do
b,tren do thi ham so y=\(\frac{-3}{2}\)x lay diem A co hoanh do la 2.tren do thi ham so y=\(\frac{2}{3}\)x.Lay diem C co hoanh do la 3.do goc AOC,sau do bieu dien diem B tren mat phang toa do sao cho OABC la hinh vuong
diem nao sau day thuoc do thi ham so y =\(-\frac{1}{3}x\)voi A(1;0) ; B(-1;-2) ; C(3;-1) ; D(1;\(\frac{1}{3}\))
mik giúp bạn giải
y =-1/3x (1)
thay x=1 vào (1) ta có : y=-1/3*1=-1/3 (loại)thay x=-1 vào (1) ta có : y= -1/3 *-1 =1/3 (loại)thay x=3 vào (1) ta có :y= -1/3*3=-1 (t/m)thay x=1 vào (1) ta có : y=-1/3*1=-1/3 (loại)Vậy điểm C thuộc đồ thị hàm số y=-1/3x
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cau 1: tinh gia tri cua x thoa manleft(x-3right)left(x^2+3x+9right)+xleft(x+2sqrt{2}right)left(2sqrt{2}-xright)-3cau 2.tinh GTLN cua bieu thuc2x-2x^2+13cau 3. tinh gia tri cua bieu thuc frac{3^{left(x+yright)^2}}{3^{left(x-yright)^2}}voi xyfrac{1}{2}cau 4. tim GTLN cua-3x^2-6x-4cau 5. cho ham so : f(x)frac{1}{5x+9}tinh gia tri cua fleft(frac{40}{25}right)cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
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cau 1: tinh gia tri cua x thoa man
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\sqrt{2}\right)\left(2\sqrt{2}-x\right)=-3\)
cau 2.tinh GTLN cua bieu thuc
\(2x-2x^2+13\)
cau 3. tinh gia tri cua bieu thuc
\(\frac{3^{\left(x+y\right)^2}}{3^{\left(x-y\right)^2}}\)voi xy=\(\frac{1}{2}\)
cau 4. tim GTLN cua
\(-3x^2-6x-4\)
cau 5. cho ham so : f(x)=\(\frac{1}{5x+9}\)
tinh gia tri cua \(f\left(\frac{40}{25}\right)\)
cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
1.tìm các nghiem nguyen cua phuong trinh: 54x^3+1=y^3
2.cho x+y=1 và xy khac 0.chung mih \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)
3.cho a,b,c la cac so thuc duong.chung minh :\(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)^2+\frac{14abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
Câu 2 thế y = 1 - x rồi quy đồng như bình thường là ra bn nhé
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cho ham so y= f(x) =6/x
tinh f(1) ; f(1.5) ;f(2) ;f(3) ;f(-2/3)
tim x biet y=3 ; y=-2
tim y biet 1<x>3 ; 1.5 be hon hoac bang x lon hon hoac bang 6
diem nao trong cac diem sau day khong thuoc do thi ham s
A(-1;-6 ) ; B(-0.5 ;-12 ) ;C(-0.5 ; -13 ) ; D(-1/3 ; -3 )
f(1)=6 ,f(2)=3,f(3)=2
b,y=3=>2
=>y=-2=>x=-3
c điểm ko thuộc đồ thị h/s là điểm
A(-1,-6)=6/-1=-6=>A THUOC H/S TREN
CÂU TIẾP THEO TƯƠNG TỰ
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a,frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:frac{10x-10y}{x^3+y^3}b,(frac{x+2}{x+1}-frac{2x}{x-1}).frac{3x+3}{x}+frac{4x^2+x+7}{x^2-x}c,frac{2}{xy}:left(frac{1}{x}-frac{1}{y}right)-frac{x^2-y^2}{left(x-yright)^2}d,frac{frac{x-y}{x+y}-frac{x+y}{x-y}}{1-frac{x^2}{x^2+y^2}}e,left(frac{1}{x+1}-frac{3}{x^3+1}+frac{3}{x^2-x+1}right).frac{3x^2-3x+3}{left(x+1right)left(x+2right)}+frac{2x-2}{x^2+2x}
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\(a,\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(b,(\frac{x+2}{x+1}-\frac{2x}{x-1}).\frac{3x+3}{x}+\frac{4x^2+x+7}{x^2-x}\)
\(c,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(d,\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)
\(e,\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}+\frac{2x-2}{x^2+2x}\)
a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)
d) \(\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)
\(=\frac{\frac{x^2-2xy+y^2}{x^2-y^2}-\frac{x^2+2xy+y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)
\(=\frac{\frac{2x^2+2y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)
\(=\frac{2x^2+2y^2}{x^2-y^2}.\frac{x^2+y^2}{y^2}\)
\(=\frac{2x^4+4x^2y^2+2y^4}{x^2y^2-y^4}\)
Xem thêm câu trả lời
13 tháng 1 2018 lúc 20:38
cho ham so y =f(x) = 4x+5
a, tinh f(-2,5); f (3) ; \(f\left(\dfrac{-3}{2}\right)\)
b , tim x khi f (x) = 2 ;f(x) =\(\dfrac{-1}{3}\) ; f(x) = 0
c, cac diem A (1;9);B (-2 ;3) ;C (3 ; 17); D(-3;7) diem nao thuoc do thi ham so
giup minh nhe minh dang can gap (luu y chi giai minh cau c ,cac cau khac minh lam roi)
c) +)Điểm A ( 1;9) => x = 1 ; y = 9
Thay x = 1 vào y = 4x+5 , ta có:
y = 4.1+5
y = 4+5
y = 9
Vậy điểm A ( 1;9 ) thuộc đồ thị hàm số y = 4x +5
+) Điểm B ( -2;3 ) => x = -2 ; y = 3
Thay x = -2 vào y = 4x +5 , ta có:
y = 4.(-2) + 5
y = (-8) + 5
y = (-3)
Vậy điểm B ( -2;3) không thuộc đồ thị hàm số y = 4x+5
....Các câu khác tương tự....> . <...
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Bleft[left(frac{x}{y}-frac{y}{x}right):left(x-yright)-2left(frac{1}{y}-frac{1}{x}right)right]:frac{x-y}{y}Cleft(frac{x+y}{2x-2y}-frac{x-y}{2x+2y}-frac{2y^2}{y-x}right):frac{2y}{x-y}D3x:left{frac{x^2-y^2}{x^3+y^3}left[left(x-frac{x^2+y^2}{y}right):left(frac{1}{x}-frac{1}{y}right)right]right}Efrac{2}{xleft(x+1right)}+frac{2}{left(x+1right)left(x+2right)}+frac{2}{left(x+2right)left(x+3right)}
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\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(E=\frac{2}{x\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
mann nào trả lời đc thui k hết 5 cái nick lun :D
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\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
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bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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tim m de ham so y=\(\frac{x^2+\left(m-1\right)x+1}{x+m-1}\) dat cuc dai tai x=2
bài 1: tính Afrac{4^2-3x+17}{x^3-1}+frac{2x-1}{x^2+x+1}+frac{6}{1-x}Bfrac{3x+1}{x^2-2x+1}-frac{1}{x+1}+frac{x+3}{1-x^2}Cleft(frac{x}{x+1}+1right):left(1-frac{3x^2}{1-x2}right)Dleft(frac{x^2}{y^2}+frac{y}{x}right):left(frac{x}{y^2}-frac{1}{y}+frac{1}{x}right)Eleft(frac{1}{x^2+4x+4}-frac{1}{x^2-4x+4}right):left(frac{1}{x+2}+frac{1}{x-2}right)Ffrac{1}{x-1}-frac{x^3-x}{x^2+1}left(frac{1}{x^2-2x+1}+frac{1}{1-x^2}right)Mấy thánh ơi giúp em với mai phải nộp rồi!!!!!!!!!!!!!!!
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bài 1: tính
\(A=\frac{4^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(B=\frac{3x+1}{x^2-2x+1}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)
\(C=\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x2}\right)\)
\(D=\left(\frac{x^2}{y^2}+\frac{y}{x}\right):\left(\frac{x}{y^2}-\frac{1}{y}+\frac{1}{x}\right)\)
\(E=\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)
\(F=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)
Mấy thánh ơi giúp em với mai phải nộp rồi!!!!!!!!!!!!!!!