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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 :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
a) \(13,5\times5,8-8,3\times4,2-5,8\times8,3+4,2\times13,5\)
\(=13,5\times\left(5,8+4,2\right)-8,3\times\left(4,2+5,8\right)\)
\(=10\times\left(13,5-8,3\right)=10\times5,2=52\)
b) \(7,8\times55,1+92,2\times55,1-7,8\times5,1-92,2\times5,1\)
\(=55,1\times\left(7,8+99,2\right)-5,1\times\left(7,8+99,2\right)\)
\(=100\times\left(55,1-5,1\right)=100\times50=5000\)
c) \(N=a^3-a^2b-ab^2+b^3=\left(a-b\right)^2\left(a+b\right)\)
\(=\left(5,75-4,25\right)^2.\left(5,75+4,25\right)=1,5^2\times10=22,5\)
Ta có M= (a3+b3)-(a2b+ab2)
=(a+b)(a2+ab+b2)-ab(a+b)
=(a+b)(a2+b2)
Thay a=5,75 ; b=4,25 vào M ta có M=(5,75+4,25)(5,752+4,252)
=511,25
\(M=a^{3^{ }}-a^2b-ab^{2^{ }}+b^3\)
\(M=a^{2^{ }}\left(a-b\right)-\left(a-b\right).b^2\)
\(M=\left(a-b\right).\left(a^2-b^2\right)\)
\(M=\left(5,75-4,25\right)^2.\left(5,75+4,25\right)\)
\(M=1,5^{2^{ }}.10\)
\(M=22,5\)
a: \(=5x^4-4x^3y+10x^3y-8x^2y^2-25x^2y^2+20xy^3-15xy^3+12y^4\)
\(=5x^4+6x^3y-33x^2y^2+5xy^3+12y^4\)
b: \(=\left(x^2-x-2\right)\left(2x-1\right)\)
\(=2x^3-x^2-2x^2+x-4x+2\)
\(=2x^3-3x^2-3x+2\)
c: \(=8x^3+y^3\)
d: \(=a^4-b^4\)
4)
a,
\(34^2+66^2+68\cdot66\\ =34^2+68\cdot66+66^2\\=34^2+2\cdot34\cdot68+66^2\\ =\left(34+66\right)^2\\ =100^2 =10000\)
b,
\(74^2+24^2-48\cdot74\\ =74^2-48\cdot74+24^2\\ =74^2-2\cdot24\cdot74+24^2\\ =\left(74-24\right)^2\\ =50^2=2500\)
c,
\(729^2-728^2\\ =\left(729+728\right)\left(729-728\right)\\ =1457\cdot1\\ =1457\)
d,
\(1001^2-1\\ =1001^2-1^2\\ =\left(1001+1\right)\left(1001-1\right)\\ =1002\cdot1000\\ =1002000\)
5)
a,
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =1\cdot\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\\ =2^{16}-1\)
b,
\(7\cdot\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{24}-1\right)\left(2^{24}+1\right)\\ =2^{48}-1\)
Bài làm
a) x3 + 12x2 + 48x + 64 tại x =6
Ta có: x3 + 12x2 + 48x + 64
<=> x3 + 3 . x2 . 4 + 3 . x . 42 + 33
<=> ( x + 3 )3
Thay x = 6 vào ( x + 3 )3 ta được:
( 6 + 3 )3
= 93 = 729
Vậy giá trị của biểu thức là 729 tại x = 6
b) x3 - 6x2 + 12x - 8 tại x = 22
Ta có: x3 - 6x2 + 12x - 8
<=> x3 - 3 . x2 . 2 + 3 . x . 22 - 23
<=> ( x - 2 )3
Thay x = 22 vào ( x - 2 )3 ta được:
( 22 - 2 )3 = 203 = 8000
Vậy giá trị của biểu thức trên là 8000 tại x = 22.
# Học tốt #
bài 1 : điền vào chỗ chấm để đk khẳng định đúng :
a) (.x..+2y...)2=x2+..4y.+4y2
b) (.a..-.3b..)2=a2-6ab+.9b2..
c) (.m..+.\(\frac{1}{2}\)..)2=.m2..+m+1/4
d) 25a2-..\(\frac{1}{4}b\).=(.5a..+1/2b)(..5a..-1/2b)
e)(.2x...+.1..)^2 = 4x^2 +.4x..+1
g)(2-x)(.4..+.2x..+.x2..)=8-x^3
h) 16a^2 - ..9. = (..4a.+3)(..4a.-3)
f)25 - ..30y.+9y^2=(..5.+...3y.)^2
Bài 5:
a: \(8A=8+8^2+...+8^8\)
\(\Leftrightarrow7A=8^8-1\)
hay \(A=\dfrac{8^8-1}{7}\)
b: \(8B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Leftrightarrow8B=3^{16}-1\)
hay \(B=\dfrac{3^{16}-1}{8}\)