bai 1 tinh
a ) (x +3) (x^2 + 3x -5)
b ) (2x -5 )^2 - (2x +5 )^2
c) (x + 5) ^2 - (x +5 ) (x- 1)
bai 2 phan tich
a) 4x^2 - 6x
b) x ^4 + 2x^3 -4x -4
c) x^3 - y^3 -4x - 4y
d) (5x -4 )^2 -49 x^2
1. Các hằng đẳng thức sau là đúng
a. x^2+6x+9/x^2+3=x+3/x+1
b. x^2-4/5x^2+13x+6=x+2/5x+3
c. x^2+5x+4/2x^2+x-3=x^2+3x+4/2x^2-5x+3
d. x^2-8x+16/16-x^2=4-x/4+x
2. P là đa thức nào để x^2+2x+1/P=x^2-1/4x^2-7x+3
a. P=4x^2+5x-2
b. P=4x^2+x-3
c. P=4x^2-x+3
d. P=4x^2+x+3
3. Đa thức Q trong đẳng thức 5(y-x)^2/5x^2-5xy=x-y/Q
a. x+y
b. 5(x+y)
c. 5(x-y)
d. x
4. Đa thức Q trong hằng đẳng x-2/2x^2+3=2x^2-4x/Q là:
a. 4x^2+16
b. 6x^2-4x
c. 4x^3+6x
d. khác
5. Phân thức 2x+1/2x-3 bằng phân thức:
a. 2x^2+x/2x-3
b. 2x^2+x/2x^2-3x
c. 2x+1/6x-9
d. Khác
Câu 5:B
Câu 4: C
Câu 3: D
Câu 2: A
Câu 1: A
BAI 1.phan tich cac da thuc sau thanh nhan tu:
a,2x^2-2xy-5x+5y
b,8x^2+4xy-2ax-ay
c,x^3-4x^2+4x
d,2xy-x^2-y^2+16
e,x^2-y^2-2yz-z^2
g,3a^2-6ab+3b^2-12c^2
BAI 2.tinh nhanh
a,37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5
b,35^2+40^2-25^2+80.35
BAI 3. Tim x biet:
a,x^3-1/9x=0
b,2x-2y-x^2+2xy-y^2=0
c,x(x-3)+x-3=0
d,x^2(x-3)+27-9x=0
BAI 4.Phan tich cac da thuc sau thanh nhan tu
a,x^2-4x+3
goi y :tach-4x=-x3xhoac tach3=-1+4
b,x^2+x-6
c,x^2-5x+6
d,x^4+4 (goi y:them va bot 4x^2)
BAI 5.Chung minh rang;
(3n+4)^2-16 chia het cho 3 voi moi so nguyen n.
BAI 6.Tinh gia tri cua bieu thuc sau:
M=a^3-a^2b-ab^2+b^3 voi a=5,75:b=4,25
BAI 7.Tim x biet:
a,x^2+x=6
b,6x^3+x^2=2x
Bài 1 câu g bạn kia làm sai mình sửa lại nhá
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
Để mình làm tiếp cho :))
Bài 2 :
Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)
\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)
\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)
\(=37,5.10-7,5.10\)
\(=10.30=300\)
Câu b : \(35^2+40^2-25^2+80.35\)
\(=\left(35^2+80.35+40^2\right)-25^2\)
\(=\left(30+45\right)^2-25^2\)
\(=75^2-25^2\)
\(=\left(75+25\right)\left(75-25\right)\)
\(=100.50=5000\)
Bài 3 :
Câu a : \(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)
Câu b : \(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)
Câu c :
\(x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2\left(x-3\right)+27-9x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)
Bài 4 :
Câu a :
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
Câu b :
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Câu c :
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
Câu d :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
Bài 1:
a) \(2x^2-2xy-5x+5y\)
\(=\left(2x^2-2xy\right)-\left(5x-5y\right)\)
\(=2x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-5\right)\)
b) \(8x^2+4xy-2ax-ay\)
\(=\left(8x^2+4xy\right)-\left(2ax+ay\right)\)
\(=4x\left(2x+y\right)-a\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x-a\right)\)
c) \(x^3-4x^2+4x\)
\(=x\left(x^2-4x+4\right)\)
\(=x\left(x-2\right)^2\)
d) \(2xy-x^2-y^2+16\)
\(=-\left[\left(x^2-2xy+y^2\right)-16\right]\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)
e) \(x^2-y^2-2yz-z^2\)
\(=-\left[\left(z^2+2yz+y^2\right)-x^2\right]\)
\(=-\left[\left(z+y\right)^2-x^2\right]\)
\(=-\left[\left(z+y+x\right)\left(z+y-x\right)\right]\)
g) \(3a^2-6ab+3b^2-12c^2\)
\(=\left(3a^2-6ab+3b^2\right)-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)
Phân tích đa thức thành các nhân tử:
a)x^2-(a+b)x+ab
b)7x^3-3xyz-21x^2+9z
c)4x+4y-x^2(x+y)
d)y^2+y-x^2+x
e)4x^2-2x-y^2-y
f)9x^2-25y^2-6x+10y
Phân tích đa thức thành nhân tử
a)(5x-4)(4x-5)-(x-3)(x-2)-(5x-4)(3x-2)
b)(5x-4)(4x-5)+(5x-1)(x+4)+3(3x-2)(4-5x)
c)(5x-4)^2+(16-25x^2)+(5x-4)(3x+2)
d)x^4-x^3-x+1
e)x^6-x^4+2x^3+2x^2
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
bai 1: cho cac da thuc
f(x)= x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-1/4x+2x-3
g(x)=5x^4-x^5+1/2x^4+x^5+x^2-4x^4-2x^3+3x^2+x^3-1/4
a, thu gon va sap xep cac da thuc tren theo luy thua giam dancua ien
b,tinh f(1);f(-1); g(1); g(-1)
c,tinh f(x)+g(x);f(x)-g(x)
bai 1: cho cac da thuc
f(x)= x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-1/4x+2x-3
g(x)=5x^4-x^5+1/2x^4+x^5+x^2-4x^4-2x^3+3x^2+x^3-1/4
a, thu gon va sap xep cac da thuc tren theo luy thua giam dancua ien
b,tinh f(1);f(-1); g(1); g(-1)
c,tinh f(x)+g(x);f(x)-g(x)
a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)
\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)
\(=7x^4-9x^3+\frac{7}{4}x-3\)
\(g\left(x\right)=5x^4-x^5+\frac{1}{2}x^2+x^5+x^2-4x^4-2x^3+3x^2+x^3-\frac{1}{4}\)
\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)
\(=x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}\)
b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)
\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)
\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)
\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)
c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)
\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)
f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)
\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)
bai 2 quy tac chuyen ve doi dau
a, 3x ²+6x+9-2x^5: 2x^4+3x:2x
b,4x ²y ²+y ³-2x-y ³+5x-3x ²y ³
c,18x+26x ² -48x ²+1x ³-5x ²-17x-x^4.x ²-4x^6
d,y ²-109y+27y+18y ²-7y ²+52y+9+4y ³-3y ².y ³+15y
tìm x biết
a) (6x-3) (2x+4) + (4x-1) (5-3x) = -21
b) 6x (3x+5) - 2x (9x-2) + (17-x) (x-1) + x (x-18) =0
c) (15-2x) (4x+1) - (13-4x) (2x-3) - (x-1) (x+2) + x2=52
d) (8x-3) (3x+2) - (4x+7) (x+4) = (2x+1) (5x-1) - 33
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21
<=> 12x2 + 24x - 6x - 12 + 20x - 12x2 - 5 + 3x = -21
<=> 41x = -21 + 12 + 5
<=> 41x = -4
<=> x = -4/41
Bài 1: Thực hiện phép tính
a) (3x-1)(9x2+3x+1)-4x(x-5)
b) (7x+2)(3-4x)-(x+3)(x2-3x+9)
c) (4x+3)(4x-3)-(2-x)(4+2x+x2)
d) (3x-8)(-5x+6)-(4x+1)(3x-2)
e) (3x-6)4x-2x(3x+5)-4x2
f) (5x-6)(6x-5)-x(3x+10)
Bài 2 : Tính
a) x(x+3)-x2=6
b) 2x(x-5)+x(-2x-1)=6
c) x (x+5)-(x+1)(x-2)=7
d)(3x+4)(6x-3)-(2x+1)(9x-2)=10
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^