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23.27. \(x^2-y^2-2x+1\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
23.25.
\(\left(x^2-4x\right)^2+\left(x-2\right)^2-10\)
\(=\left(x^2-4x\right)^2-4+\left(x-2\right)^2-6\)
\(=\left(x^2-4x+4\right)\left(x^2-4x-4\right)+x^2-4x+4-6\)
\(=\left(x^2-4x+4\right)\left(x^2-4x-10\right)\)
23.23
\(x^3-2x^2-6x+27\)
\(=\left(x^3+27\right)-2x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+9-2x\right)\)
\(=\left(x+3\right)\left(x^2-5x+9\right)\)
Đề số 3.
1.
a,\(4x\left(5x^2-2x+3\right)\)
\(=20x^3-8x^2+12x\)
b.\(\left(x-2\right)\left(x^2-3x+5\right)\)
\(=x^3-3x^2+5x-2x^2+6x-10\)
\(=x^3-5x^2+11x-10\)
c,\(\left(10x^4-5x^3+3x^2\right):5x^2\)
\(=2x^2-x+\dfrac{3}{5}\)
d,\(\left(x^2-12xy+36y^2\right):\left(x-6y\right)\)
\(=\left(x-6y\right)^2:\left(x-6y\right)\)
\(=x-6y\)
2.
a,\(x^2+5x+5xy+25y\)
\(=\left(x^2+5x\right)+\left(5xy+25y\right)\)
\(=x\left(x+5\right)+5y\left(x+5\right)\)
\(=\left(x+5y\right)\left(x+5\right)\)
b,\(x^2-y^2+14x+49\)
\(=\left(x^2+14x+49\right)-y^2\)
\(=\left(x+7\right)^2-y^2\)
\(=\left(x+7-y\right)\left(x+7+y\right)\)
c,\(x^2-24x-25\)
\(=x^2+25x-x-25\)
\(=\left(x^2-x\right)+\left(25x-25\right)\)
\(=x\left(x-1\right)+25\left(x-1\right)\)
\(=\left(x+25\right)\left(x-1\right)\)
3.
a,\(5x\left(x-3\right)-x+3=0\)
\(5x\left(x-3\right)-\left(x-3\right)=0\)
\(\left(5x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{5}\) hoặc \(x=3\)
b.\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(3x^2-15x-\left(2x+3x^2-2-3x\right)=30\)
\(3x^2-15x-2x-3x^2+2+3x=30\)
\(-14x+2=30\)
\(-14x=28\)
\(x=-2\)
c,\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\)
\(x^2+5x+6-x^2-5x+2x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
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uk đi đi cho đỡ tốn diện tích khi Nam đăg câu hỏi câu trả lời của Nam
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}
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)
giải chi tiết hộ mk với. mơn <3
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
\(b,9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\)
\(\Leftrightarrow9x^2-4-12x^2+23x-10=0\)
\(\Leftrightarrow-3x^2+23x-14=0\)
\(\Leftrightarrow-3x^2+21x+2x-14=0\)
\(\Leftrightarrow-3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,4x^2\left(x-1\right)-x+1=0\)
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x=-1\\x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\) \(d,x^2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-3\end{matrix}\right.\)