Bài 15:

1: \(A=4x-x^2+1\)

\(=-\left(x^2-4x-1\right)\)

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5< =5\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

2: \(B=3-4x-x^2\)

\(=-\left(x^2+4x-3\right)\)

\(=-\left(x^2+4x+4-7\right)\)

\(=-\left(x+2\right)^2+7< =7\forall x\)

Dấu '=' xảy ra khi x+2=0

=>x=-2

3: \(C=8-x^2-5x\)

\(=-\left(x^2+5x-8\right)\)

\(=-\left(x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{57}{4}\right)\)

\(=-\left(x+\dfrac{5}{2}\right)^2+\dfrac{57}{4}< =\dfrac{57}{4}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{5}{2}=0\)

=>\(x=-\dfrac{5}{2}\)

4: \(D=-x^2+6x-4\)

\(=-\left(x^2-6x+4\right)\)

\(=-\left(x^2-6x+9-5\right)\)

\(=-\left(x-3\right)^2+5< =5\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

5: \(E=-10-x^2-6x\)

\(=-\left(x^2+6x+10\right)=-\left(x^2+6x+9+1\right)\)

\(=-\left(x+3\right)^2-1< =-1\forall x\)

Dấu '=' xảy ra khi x+3=0

=>x=-3

6: \(F=-x^2+13x+1\)
\(=-\left(x^2-13x-1\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{13}{2}+\dfrac{169}{4}-\dfrac{173}{4}\right)\)

\(=-\left(x-\dfrac{13}{2}\right)^2+\dfrac{173}{4}\le\dfrac{173}{4}\forall x\)

Dấu '=' xảy ra khi x-13/2=0

=>x=13/2

7: \(G=-4x^2+8x-7\)

\(=-\left(4x^2-8x+7\right)\)

\(=-\left(4x^2-8x+4+3\right)\)

\(=-\left(2x-2\right)^2-3< =-3\forall x\)

Dấu '=' xảy ra khi 2x-2=0

=>2x=2

=>x=1

8: \(H=-4x^2-12x\)

\(=-\left(4x^2+12x\right)\)

\(=-\left(4x^2+12x+9-9\right)\)

\(=-\left(2x+3\right)^2+9< =9\forall x\)

Dấu '=' xảy ra khi 2x+3=0

=>x=-3/2

9: \(I=3x-9x^2-1\)

\(=-9\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

\(=-9\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{1}{12}\right)\)

\(=-9\left(x-\dfrac{1}{6}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

10: \(K=7-9x^2-8x\)

\(=-9\left(x^2+\dfrac{8}{9}x-\dfrac{7}{9}\right)\)

\(=-9\left(x^2+2\cdot x\cdot\dfrac{4}{9}+\dfrac{16}{81}-\dfrac{79}{81}\right)=-9\left(x+\dfrac{4}{9}\right)^2+\dfrac{79}{9}< =\dfrac{79}{9}\forall x\)

Dấu '=' xảy ra khi x+4/9=0

=>x=-4/9

ĐKXĐ: \(x\notin\left\{-1;-4;-6;3\right\}\)

\(\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\\ \Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}-\dfrac{9}{\left(x-3\right)\left(x+6\right)}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}-\left(\dfrac{1}{x-3}-\dfrac{1}{x+6}\right)=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{x-3-x-1}{\left(x+1\right)\left(x-3\right)}=\dfrac{4}{3}\\ \Rightarrow4\left(x^2-2x-3\right)=3.\left(-4\right)\\ \Leftrightarrow4x^2-8x-12=-12\\ \Leftrightarrow4x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

a) $\left(\frac12 x^5-3x^2\right):x^2-x\left(\frac12x^2+5\right)+2x(x-2)(x+2)-(x^3-27):(x+3)$

$=\frac12x^5:x^2-3x^2:x^2-\frac12x^3-5x+2x(x^2-4)-(x-3)(x^2+3x+9):(x-3)$

$=\frac12x^3-3-\frac12x^3-5x+2x^3-8x-(x^2+3x+9)$

$=2x^3-13x-3-x^2-3x-9$

$=2x^3-x^2-16x-12$

\(\left(1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3+10^3\right)\)

\(=\left(1+2+3+...+10\right)^2\)

\(=\left(\dfrac{10\cdot11}{2}\right)^2=\left(5\cdot11\right)^2=25\cdot121⋮11\)

Ta sẽ chứng minh \(1^3+2^3+3^3+...+n^3=\left[\dfrac{n\left(n+1\right)}{2}\right]^2\) bằng quy nạp.   (*)

Thật vậy, với \(n=1\) thì (*) thành \(1^3=\left[\dfrac{1.2}{2}\right]^2\), luôn đúng

Giả sử (*) đúng đến \(n=k\ge1\), khi đó cần chứng minh (*) đúng với \(n=k+1\). Thật vậy, với \(n=k+1\) thì

\(VT=1^3+2^3+3^2+...+k^3+\left(k+1\right)^3\)

\(=\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3\) (theo giả thiết quy nạp)

\(=\left(k+1\right)^2\left(\dfrac{k^2}{4}+k+1\right)\)

\(=\left(k+1\right)^2\left(\dfrac{k^2+4k+4}{4}\right)\)

\(=\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}\)

\(=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)

Vậy (*) đúng với \(n=k+1\). Theo nguyên lí quy nạp, (*) được chứng minh.

Như vậy \(1^3+2^3+3^3+...+10^3=\left(\dfrac{10.11}{2}\right)^2=\left(5.11\right)^2=25.11^2⋮11\), ta có đpcm.

Gọi mẫu số là x

(ĐIều kiện: \(x\ne0\))

Vì phân số nhỏ hơn 1 nên mẫu số>tử số

=>Mẫu số>32/2=16

Tử số là 32-x

Mẫu số khi tăng thêm 10 đơn vị là x+10

Tử số khi giảm đi một nửa là \(\dfrac{32-x}{2}\)

Phân số mới là 2/17 nên \(\dfrac{32-x}{2}:\left(x+10\right)=\dfrac{2}{17}\)

=>\(\dfrac{32-x}{2x+20}=\dfrac{2}{17}\)

=>17(32-x)=2(2x+20)

=>544-17x=4x+40

=>-21x=40-544=-504

=>x=24

Tử số là 32-24=8

Vậy: Phân số cần tìm là \(\dfrac{8}{24}\)

\(4xy^2\cdot x-\left(-12x^2y^2\right)\)

\(=4x^2y^2+12x^2y^2\)

\(=16x^2y^2\)

hơi khó nha

a)

\(2^{2024}=2^{8.11.23}\)

\(2^8\equiv4\left(mod7\right)\)

\(2^{8.11}\equiv\left(2^8\right)^{11}\left(mod7\right)\equiv4^{11}\left(mod7\right)\equiv2\left(mod7\right)\)

\(\Rightarrow2^{8.11.23}\equiv\left(2^{8.11}\right)^{23}\left(mod7\right)\equiv2^{23}\left(mod7\right)\equiv4\left(mod7\right)\)

\(\Rightarrow2^{2024}\) chia 7 dư 4

\(41^{2023}=41.\left(41^2\right)^{1011}\)

\(41^2\equiv1\left(mod7\right)\)

\(\Rightarrow\left(41^2\right)^{1011}\equiv1^{1011}\left(mod7\right)\equiv1\left(mod7\right)\)

\(\Rightarrow41.\left(41^2\right)^{1011}\equiv41.1\left(mod7\right)\equiv6\left(mod7\right)\)

\(\Rightarrow2^{2024}+41^{2023}\equiv4+6\left(mod7\right)\equiv3\left(mod7\right)\)

Vậy \(2^{2024}+41^{2023}\) chia 7 dư 3

$16(x-1)^2-25=0$

$\Leftrightarrow (4x-4)^2-5^2=0$

$\Leftrightarrow (4x-4-5)(4x-4+5)=0$

$\Leftrightarrow (4x-9)(4x+1)=0$

$\Leftrightarrow \left[\begin{array}{} 4x-9=0\\4x+1=0 \end{array} \right. \Leftrightarrow \left[\begin{array}{} 4x=9\\4x=-1 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=\frac94\\x=-\frac14 \end{array} \right.$

#$\mathtt{Toru}$

\(16\left(x-1\right)^2-25=0\)

\(16\left(x-1\right)^2=0+25\)

\(16\left(x-1\right)^2=25\)

\(\left(x-1\right)^2=\dfrac{25}{16}\)

\(x-1=\dfrac{5}{4};x-1=-\dfrac{5}{4}\)

*) \(x-1=\dfrac{5}{4}\)

\(x=\dfrac{5}{4}+1\)

\(x=\dfrac{9}{4}\)

*) \(x-1=-\dfrac{5}{4}\)

\(x=-\dfrac{5}{4}+1\)

\(x=-\dfrac{1}{4}\)

Vậy \(x=-\dfrac{1}{4};x=\dfrac{9}{4}\)

$(2x-5)^2-x^2=0$

$\Leftrightarrow (2x-5-x)(2x-5+x)=0$

$\Leftrightarrow (x-5)(3x-5)=0$

$\Leftrightarrow \left[\begin{array}{} x-5=0\\ 3x-5=0 \end{array} \right.$

$\Leftrightarrow \left[\begin{array}{} x=5\\x=\frac53 \end{array} \right.$