Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)
b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(18^8-\left(18^8-1\right)=1\)
\(c,100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)
áp dụng công thức Gauss ta đc đáp án là:10100
d, mk khỏi ghi đề dài dòng:
\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:
\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)
\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)
\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)
1c,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)
P/s : Phá ngoặc ra là ok :
a )
\(\left[4x-2\left(x-3\right)\right].\left(-3x\right)\)
\(=\left[4x-2x+6\right]\left(-3x\right)\)
\(=-12x^2+6x^2-18x\)
b )
\(3\left[x-3\left(4-2x\right)+8\right]\)
\(=3\left[x-12+6x+8\right]\)
\(=3\left[7x-4\right]\)
\(=21x-12\)
c )
\(5\left(3x^2-4y^3\right)+9\left(2x^2-y^3\right)\)
\(=15x^2-20y^3+18x^2-9y^3\)
\(=33x^2-29y^3\)
d )
\(3x^2\left(2y-1\right)-2x^2\left(5y-3\right)\)
\(=6x^2y-3x^2-10x^2y+6x^2\)
\(=-4x^2y+3x^2\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
a) 1,62 + 4 . 0,8 . 3,4 + 3,42 = 1,62 + 2 . 1,6 . 3,4 + 3,42
= (1,6 + 3,4 )2 = 52 = 25
b) 34 .54 - (152 + 1)(152 - 1) = 34.54 - ( 154 - 1)
= 34.54 - 34.54 + 1
= 1
c) Do x= 11 \(\Rightarrow\) 12 = x + 1
Thay 12 = x+1 vào biểu thức, ta có :
x4 - (x+1)x3 + (x+1)x2 - (x+1)x + 111
= x4 - x4 - x3 + x3 +x2 -x2 - x+ 111
= - x + 111 = -11 + 111 = 100 ( do x = 11)
Vậy ................................
_______________JK ~ Liên Quân Group ______________
a)
\((x+2)(x+4)(x+6)(x+8)+16\)
\(=[(x+2)(x+8)][(x+4)(x+6)]+16\)
\(=(x^2+10x+16)(x^2+10x+24)+16\)
\(=a(a+8)+16\) (Đặt \(x^2+10x+16=a\) )
\(=a^2+2.4.a+4^2=(a+4)^2\)
\(=(x^2+10x+16+4)^2\)
\(=(x^2+10x+20)^2\)
b) \((x^2+x)(x^2+x+1)-6\)
\(=(x^2+x)^2+(x^2+x)-6\)
\(=(x^2+x)^2-2(x^2+x)+3(x^2+x)-6\)
\(=(x^2+x)(x^2+x-2)+3(x^2+x-2)\)
\(=(x^2+x-2)(x^2+x+3)\)
\(=(x^2-x+2x-2)(x^2+x+3)\)
\(=[x(x-1)+2(x-1)](x^2+x+3)\)
\(=(x-1)(x+2)(x^2+x+3)\)
c)
\((x^2-4x)^2-8(x^2-4x)+15\)
\(=(x^2-4x)^2-3(x^2-4x)-5(x^2-4x)+15\)
\(=(x^2-4x)(x^2-4x-3)-5(x^2-4x-3)\)
\(=(x^2-4x-3)(x^2-4x-5)\)
\(=(x^2-4x-3)(x^2+x-5x-5)\)
\(=(x^2-4x-3)[x(x+1)-5(x+1)]=(x^2-4x-3)(x+1)(x-5)\)
a)\(\left(x^4+8x^2+16\right):\left(x^2+4\right)\)
\(=\left(x^2+4\right)^2:\left(x^2+4\right)\)
\(=x^2+4\)
b)\(\left(25-x^2\right):\left(x+5\right)\)
=\(\left(x^2-5^2\right):\left(x+5\right)\)
\(=\left(x-5\right)\left(x+5\right):\left(x+5\right)\)
\(=x-5\)
c)\(\left(x^3+1\right):\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right):\left(x^2-x+1\right)\)
\(=x+1\)
a) \(\left(x^4+8x^2+16\right):\left(x^2+4\right)\)\(=\left(x^2+4\right)^2:\left(x^2+4\right)\)\(=x^2+4\)
b) \(\left(25-x^2\right):\left(x+5\right)=\left(x-5\right).\left(x+5\right):\left(x+5\right)\)\(=x-5\)
c) \(=\left(x^3+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\)\(=x+1\)
Học tốt
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
a)3
b)101018000
c)4