Ôn tập cuối năm phần số học - Hỏi đáp

11=11

11²=121

11³=1331

11⁴=14641

.....

11¹⁰=1....01

=>11¹⁰-1=1...01-1=1...00

=>11¹⁰-1 \(⋮\)100

Câu a và câu c bn kia làm rồi nên mk làm câu b thôi nhé....

b) y² + 4^x + 2y - 2^x+1 + 2 = 0

\(\Leftrightarrow\) (y² + 2y + 1) + 4^x - 2^x.2 + 1 = 0

\(\Leftrightarrow\) (y + 1)² + [(2^x)² - 2.2^x.1 + 1] = 0

\(\Leftrightarrow\) (y + 1)² + (2^x - 1)² = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\2^x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)

Vậy...................

ảnh k đc rõ mấy, mong bạn thông cảm :)

\(\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3+\sqrt{5}}\right)}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3-\sqrt{5}}\right)}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5+2\sqrt{5}+1}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5-2\sqrt{5}+1}}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{\left(\sqrt{5}-1\right)^2}}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5}-1}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}\)

=\(\dfrac{(3\sqrt{2}+\sqrt{10})\left(3\sqrt{5}-1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}-\dfrac{(3\sqrt{2}-\sqrt{10})\left(3\sqrt{5}+1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)

=\(\dfrac{9\sqrt{10}-3\sqrt{2}+3\sqrt{50}-\sqrt{10}-\left(9\sqrt{10}+3\sqrt{2}-3\sqrt{50}-\sqrt{10}\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)

=\(\dfrac{-6\sqrt{2}+6\sqrt{50}}{\left(3\sqrt{5}\right)^2-1}\)=\(\dfrac{-6\left(\sqrt{2}-\sqrt{50}\right)}{45-1}\)

=\(\dfrac{-6\sqrt{2}\left(1-5\right)}{44}=\dfrac{24\sqrt{2}}{44}=\dfrac{6\sqrt{2}}{11}\)

Sau hai giờ 2 xe cách nhau 100km

Ta có: \(\dfrac{16}{2x+y+z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Leftrightarrow\dfrac{1}{2x+y+z}\le\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(1\right)\)

Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\left(2\right)\\\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\left(3\right)\end{matrix}\right.\)

Cộng (1), (2), (3) vế theo vế ta được:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{4}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{4.4}{16}=1\)

Dấu = xảy ra khi \(x=y=z=\dfrac{3}{4}\)

a) x³ - 7x + 6

= x³ - 7x + 7 - 1

= ( x - 1)( x² + x + 1) - 7( x - 1)

= ( x - 1)( x² + x - 6)

= ( x - 1)( x² - 2x + 3x - 6)

= ( x - 1)[ x( x - 2) + 3( x - 2)]

= ( x - 1)( x - 2)( x + 3)

b) ab( a + b) - bc( b+ c) + ac( a - c)

= a²b + ab² - b²c - bc² + ac( a - c)

= b(a² - c²) + b²( a - c) + ac( a - c)

= b( a + c)( a - c) + b²( a - c) + ac( a - c)

= ( a - c)( ab + bc + b² + ac)

= ( a - c) [ b( a + b) + c( a + b) ]

= ( a - c)( a + b)( b + c)

E xin ủng hộ cách khác cho bài 2 :(

Áp dụng hđt mở rộng ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=ab+bc+ac\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

\(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Với \(a=b=c\) ta có: \(N=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\) ta có: \(N=\dfrac{-abc}{abc}=-1\)

bài 1

A=

\(\dfrac{a}{ab+a+2006}+\dfrac{b}{bc+b+1}+\dfrac{2006c}{ac+2006c+2006}\)

=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc.c}{ac+abc.c+abc}\)=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc.c}{c\left(a+abc+ab\right)}\)=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc}{a+abc+ab}\)=\(\dfrac{a+ab+abc}{ab+a+abc}=1\)

Vậy A=1

Sửa đề: thì \(ay-bx=0\)

Giải:

Xét hiệu: \(\left(a^2+b^2\right)\left(x^2+y^2\right)-\left(ax+by\right)^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-b^2y^2\) \(-2axby\)

\(=a^2y^2-2axby+b^2x^2\)

\(=\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\) (Đpcm)

Ta có: (a² + b²)(x² + y²) = (ax + by)²

<=> a²x² + a²y² + b²x² + b²y² = a²x² + 2axby + b²y²

<=> a²x² + a²y² + b²x² + b²y² - a²x² - 2axby - b²y² = 0

<=> (a²y² - axby) + (b²x² - axby) = 0

<=> ay(ay - bx) - bx(ay - bx) = 0

<=> (ay - bx)² = 0

<=> ay - bx = 0

Vậy bài toán đã được chứng minh

gọi số đó là : 10a+b

ta có : \(\sqrt{10a+b}\)= a+\(\sqrt{b}\)

Để \(\sqrt{10a+b}\) nguyên thì \(\sqrt{b}\) nguyên \(\Leftrightarrow\)

b\(\in\left\{0;1;4;9\right\}\)

ta có : ( \(\sqrt{10a+b}\))²=a²+b +2a.\(\sqrt{b}\)

\(\Rightarrow\) 10a+b=a²+b+2a.\(\sqrt{b}\)

\(\Rightarrow\)a(a-10+2\(\sqrt{b}\))=0

\(\Rightarrow\)\(\left[{}\begin{matrix}a=0\left(loai\right)\\a+2\sqrt{b}-10=0\end{matrix}\right.\)

Th2 : a+2.\(\sqrt{b}\)-10 = 0\(\Rightarrow\) a=10-2.\(\sqrt{b}\).Xét tất cả các trường hợp b=1;4;9 thì tìm được các giá trị thỏa mãn là a=8;6 ; 4

Lời giải:

Vì \(x^2-2019x+1=0\Rightarrow x^2+1=2019x\)

\(A=\frac{x^4+x^2+1}{x^2}=\frac{x^4+2x^2+1-x^2}{x^2}=\frac{(x^2+1)^2-x^2}{x^2}\)

\(=\frac{(2019x)^2-x^2}{x^2}=\frac{x^2(2019^2-1)}{x^2}=2019^2-1\)

Vậy \(A=2019^2-1\)

Sửa đề tí nhé :P

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)

\(=-\left(a^2-2ab+b^2-c^2\right)\left(a^2+2ab+b^2-c^2\right)\)

\(=\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]\)

\(=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)

Có lộn đề không ( a² + b² + c² )²

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