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đề 1 bài 4
xét tam gics ABC và tam giác HBA có
góc B chung
góc BAC = góc BHA (=90 độ)
=> tam giác ABC đồng dạng vs tam giác HBA (g.g)
=> AB/HB=BC/AB=> AB^2=HB *BC
áp dụng đl py ta go trog tam giác vuông ABC có
BC^2 = AB^2 +AC^2=6^2+8^2=100
=> BC =\(\sqrt{100}\)=10 cm
ta có tam giác ABC đồng dạng vs tam giác HBA (cm câu a )
=> AC/AH=BC/BA=>AH=8*6/10=4.8CM
=>AB/BH=AC/AH=> BH=6*4.8/8=3,6cm
=>HC =BC-BH=10-3,6=6,4cm
dề 1 bài 1
5x+12=3x -14
<=>5x-3x=-14-12
<=>2x=-26
<=> x=-12
vạy S={-12}
(4x-2)*(3x+4)=0
<=>4x-2=0<=>x=1/2
<=>3x+4=0<=>x=-4/3
vậy S={1/2;-4/3}
đkxđ : x\(\ne2;x\ne-3\)
\(\dfrac{4}{x-2}+\dfrac{1}{x+3}=0\)
<=> 4(x+3)/(x-2)(x+3)+1(x-2)/(x-2)(x+3)
=> 4x+12+x-2=0
<=>5x=-10
<=>x=-2 (nhận)
vậy S={-2}
\(B=\sqrt{371^2}+2\sqrt{31^2}-\sqrt{121^2}=371+2.31-121=371+62-121=312\)
Câu 1 : Làm tính nhân :
a) \(2x\left(x^2-7x-3\right)\)
\(=2x^3-14x-6x\)
b) \(\left(-2x^3+3y^2-7xy\right).4xy^2\)
\(=-8x^4y^2+3x-28x^2y^3\)
c) \(\left(25x^2+10xy+4y^2\right).\left(5x-2y\right)\)
\(=-50x^2y-20xy^2-8y^3+125x^3+50x^2y+20xy^2\)
\(=-8y^3+125x^3\)
d) \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=10x^3-2x^2+4x-6-5x^4+x^3-2x^2+3x+20x^5-4x^4+8x^3-12x^2\)
\(=20x^5-9x^4+19x^3-16x^2-7x-6\)
Câu 3: phân tích
a)\(4x-8y\)
\(=4\left(x-2y\right)\)
b)\(x^2+2xy+y^2-16\)
\(=\left(x+y\right)^2-4^2\)
\(=\left(x+y-4\right)\left(x+y+4\right)\)
c)\(3x^2+5x-3xy-5y\)
\(=3x^2-3xy+5x-5y\)
\(=3x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+5\right)\)
bài 4
a)xy+y2-x-y
=(xy+y2)-(x+y)
=y(x+y)-(x+y)
=(x+y)(y-1)
b)25-x2+4xy-4y2
=25-(x2-4xy+4y2)
=25-(x-2y)2
=[5-(x-2y)][5+(x-2y)]
=(5-x+2y)(5+x-2y)
c) xy+xz-2y-2z
=(xy+xz)-(2y+2z)
=x(y+z)-2(y+z)
=(y+z)(x-2)
Bài 7: Cứng minh đẳng thức
b) \(\left(x^{n+3}-x^{n+1}.y^2\right)\div\left(x+y\right)=x^{n+2}-x^{n+1}.y\)
Biến đổi vế trái
\(\left(x^{n+3}-x^{n+1}.y^2\right)\div\left(x+y\right)\)
\(=\left(x^n.x^3-x^n.x.y^2\right)\div\left(x+y\right)\)
\(=x^n.x\left(x^2-y^2\right)\div\left(x+y\right)\)
\(=x^{n+1}\left(x-y\right)\left(x+y\right)\div\left(x+y\right)\)
\(=x^{n+1}\left(x-y\right)\)
Biến đổi vế phải
\(x^{n+2}-x^{n+1}.y\)
\(=x^n.x^2-x^n.x.y\)
\(=x^n.x\left(x-y\right)\)
\(=x^{n+1}\left(x-y\right)\) bằng vế trái (điều phải chứng minh)
Câu 3 ( Đề 1)
a) A = ( x - 2)2 - ( x + 3)( x - 3)
A = x2 - 4x + 4 - x2 + 9
A = - 4x + 13
b) B = 4x( x + 3) - 3x(4 + x)
B = 4x2 + 12x - 12x - 3x2
B = x2
Câu 4 . a) 5x3 - 45x
= 5x( x2 - 32)
= 5x( x - 3)( x + 3)
b) 5x2 + 5xy - x - y
= 5x( x + y) - ( x +y)
= ( x + y)( 5x - 1)
c) x3 - 9x2y + xy2 - 9y3
= x( x2 + y2) - 9y( x2 + y2)
= ( x2 + y2)( x - 9y)
Câu 3 : ( đề 2)
a) A = ( x - 2)2 -( x + 1)( x - 1) - x( 1 - x)
A= x2 - 4x + 4 - x2 + 1 - x + x2
A = x2 - 5x + 5
b)B = 7x( x - 4) - 2x( x - 6)
B = 7x2 - 28x - 2x2 + 12x
B = 5x2 - 16x
Cau 4 .
a) 4x3 - 64x
= 4x( x2 - 42)
= 4x( x - 4)( x + 4)
b) x3 + x + 5x2 + 5
= x( x2 + 1) + 5( x2 + 1)
= ( x2 + 1)( x + 5)
c) x2 - 3xy - 10y2
= x2 - (2y)2 - 3xy - 6y2
= ( x - 2y)( x + 2y) - 3y( x + 2y)
= ( x + 2y)( x - 5y)
Cau 5 . 4x2 - 5x + x3 - 20
= x2( x + 4) - 5( x + 4)
= ( x + 4)( x2 - 5)
Vay phep chia : ( 4x2 - 5x + x3 - 20) cho da thuc ( x + 4) duoc thuong la x2 - 5
bài 4
a) 4x3-64x
= 4x(x2-16)
b)x3+x+5x2+5
= (x3+x)+(5x2+5)
= x(x2+1)+5(x2+1)
= (x2+1)(x+5)
a) x3 - 4x2 + 4x
= x(x2 - 4x + 4)
= x(x - 2)2
b) x2 - 3x + 2
= x2 - x - 2x + 2
= (x2 - x) + (2x - 2)
= x(x - 1) + 2(x - 1)
= (x + 2)(x - 1)
c) 8x3 + \(\dfrac{1}{27}\)
= \(\left(2x+\dfrac{1}{3}\right)\)\(\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)
d) 64x3 - \(\dfrac{1}{8}\)
= \(\left(4x+\dfrac{1}{2}\right)\left(16x^2-2x+\dfrac{1}{4}\right)\)
e) x2 - 4 + (x - 2)2
= (x + 2)(x - 2) - (x - 2)2
= (x - 2)[(x + 2) - (x - 2)]
= (x - 2)(x + 2 - x + 2)
= 4(x - 2)
f) x3 - 2x3 + x - xy2
= -x3 + x - xy2
= -x(x2 - 1 + y2)
g) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x + 3)(x2 - 3x + 9) - 4x(x + 3)
= (x + 3)[(x2 - 3x + 9) - 4x]
= (x + 3)(x2 - 3x + 9 - 4x)
= (x + 3)(x2 - 7x + 9)
h) 2x - 2y - x2 + 2xy - y2
= (2x - 2y) - (x2 - 2xy + y2)
= 2(x - y) - (x - y)2
= (x - y)(2 - x + y)
i) 3x2 + 6x + 3 - 3y2
= 3(x2 + 2x + 1 - y2)
= 3[(x2 + 2x + 1) - y2]
= 3[(x + 1)2 - y2]
= 3( x + 1 - y)(x + 1 + y)
k) 25 - x2 - y2 + 2xy
= 25 - (x2 - 2xy + y2)
= 25 - (x - y)2
= (5 - x + y)(5 + x - y)
l) 3x - 3y - x2 + 2xy - y2
= (3x - 3y) - (x2 - 2xy + y2)
= 3(x - y) - (x - y)2
= (x - y)(3 - x + y)
m) x2 - y2 + 2x - 2y
= (x2 - y2) + (2x - 2y)
= (x - y)(x + y) + 2(x - y)
= (x - y)(x + y + 2)
n) x4 + 2x3 - 4x - 4
= (x4 - 4) + (2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
o) x2(1 - x2) - 4x - 4x2
= x2(1 - x)( 1 + x) - 4x(1 + x)
= x(1 + x)[x(1 - x) - 4x]
= x(x + 1)(x - x2 - 4)
p) x3 + y3 + z3 - 3xyz
= x3 + y3 + z3 - 3x2y + 3x2y - 3xy2 + 3xy2 - 3xyz
= [(x3 + 3x2y + 3xy2 + y3) + z3] - (3x2y + 3xy2 + 3xyz)
= [(x + y)3 + z3] - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - (x + y)z + z2] - 3xy(x + y + z)
= (x + y + z)(x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - xz - yz)
q) (x - y)3 + (y - z)3 + (z - x)3
= [(x - y) + (y - z)][(x - y)2 - (x - y)(y - z) + (y - z)2] + (z - x)3
= (x - z)(x2 - 2xy + y2 - xy + xz - y2 + yz + y2 - 2yz + z2) - (x - z)3
= (x - z)(x2 + y2 + z2 - 3xy + xz - yz) - (x - z)3
= (x - z)[x2 + y2 + z2 - 3xy + xz - yz - (x - z)2]
= (x - z)(x2 + y2 - 3xy + xz - yz - x2 + 2xz - z2)
= (x - z)(y2 - 3xy + 3xz - yz)
= (x - z)[(y2 - yz) - (3xy - 3xz)]
= (x - z)[y(y - z) - 3x(y - z)
= (x - z)(y - 3x)(y - z)
Nhớ tik nha
a)\(2x^2-7xy+5y^2\)
\(=2x^2-2xy-5xy+5y^2\)
\(=2x\left(x-y\right)-5y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-5y\right)\)
b)\(x^3+3x^2y-4xy^2-12y^3\)
\(=\left(x^3+3x^2y\right)-\left(4xy^2+12y^3\right)\)
\(=x^2\left(x+3y\right)-4y^2\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x^2-4y^2\right)\)
\(=\left(x+3y\right)\left(x-2y\right)\left(x+2y\right)\)
a)Thay x=1 ta có:
1+m.1-4.1-4=0
<=>m-7=0
<=>m=7
b)Với m=7 ta có:
x3+7x2-4x-4=0
<=>(x3-x2)+(8x2-8x)+(4x-4)=0
<=>(x-1)(x2+8x+4)=0
=>x2+8x+4=0
<=>x2+8x+16-12=0
<=>(x+4)2=12
<=>x+4=\(^+_-\sqrt{12}\)
<=>x=\(\sqrt{12}\)-4 hoặc x=\(-\sqrt{12}-4\)
Vậy...
Trả lời:
\(\sqrt{4x^2}=\sqrt{\left(2x\right)^2}=\left|2x\right|\)
+) TH1: Nếu \(x\ge0\Rightarrow\left|2x\right|=2x\)
+) TH2: Nếu \(x< 0\Rightarrow\left|2x\right|=-2x\)