Nguyễn Thanh Hằng

\(2x=3y=4z\)

\(\Leftrightarrow\dfrac{2x}{12}=\dfrac{3y}{12}=\dfrac{4z}{12}\)

\(\Leftrightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}\)

Đặt :

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=4k\\z=3k\end{matrix}\right.\)

\(2x^2-3z^2=1125\Leftrightarrow2.\left(6k\right)^2-3.\left(3k\right)^2=1125\Leftrightarrow72k^2-27k^2=1125\)

\(\Leftrightarrow45k^2=1125\)

\(\Leftrightarrow k^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)

Với \(k=5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.5=30\\y=4.5=20\\z=3.5=15\end{matrix}\right.\)

Với \(k=-5\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6.\left(-5\right)=-30\\y=4.\left(-5\right)=-20\\z=3.\left(-5\right)=-15\end{matrix}\right.\)

Vậy ...

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}\)

\(=\dfrac{ab+ac+bc+ba-ca-cb}{2+3-4}=\dfrac{2ab}{1}\) \(\left(1\right)\)

\(=\dfrac{bc+cb+bc+ba-ab-ac}{3+4-2}=\dfrac{2bc}{5}\left(2\right)\)

\(=\dfrac{ab+ac+ca+cb-bc-ba}{2+4-3}=\dfrac{2ac}{3}\)\(\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow\dfrac{2ab}{1}=\dfrac{2bc}{5}=\dfrac{2ac}{3}\)

\(\dfrac{2ab}{1}=\dfrac{2bc}{5}\Leftrightarrow\dfrac{a}{1}=\dfrac{c}{15}\) \(\Leftrightarrow\dfrac{a}{3}=\dfrac{c}{15}\left(I\right)\)

\(\dfrac{2bc}{5}=\dfrac{2ac}{3}\Leftrightarrow\dfrac{b}{5}=\dfrac{a}{3}\left(II\right)\)

Từ \(\left(I\right)+\left(II\right)\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\left(đpcm\right)\)

\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}\)

\(\Leftrightarrow5\left(y^2-x^2\right)=3\left(y^2+x^2\right)\)

\(\Leftrightarrow5y^2-5x^2=3y^2+3x^2\)

\(\Leftrightarrow2y^2=8x^2\)

\(\Leftrightarrow y^2=4x^2\)

\(\Leftrightarrow y^{10}=1024.x^{10}\)

Mà \(x^{10}.y^{10}=1024\)

\(\Leftrightarrow x^{10}.1024x^{10}=1024\)

\(\Leftrightarrow x^{20}=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

+)Với \(x=1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

+) Với \(x=-1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy...

1/ a/ \(\left(x+y\right)^3=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)\left(x^2+2xy+b^2\right)=x^3+2x^2y+x^2y+xy^2+2xy^2+y^3=x^3+3x^2y+3xy^2+y^3\)

b/ \(\left(x-y\right)^3=\left(x-y\right)\left(x-y\right)^2=\left(x-y\right)\left(x^2-2xy+y^2\right)=x^3-2x^2y-x^2y+2xy^2+xy^2-y^3=x^3-3x^2y+3xy^2+y^3\)2/

a/ \(x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow x=6\)

Vậy ...

b/ \(\left(x-1\right)^3-x\left(x^2-3x+1\right)=18\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+3x^2-x=18\)

\(\Leftrightarrow2x-1=18\)

\(\Leftrightarrow x=\dfrac{19}{2}\)

Vậy...

3/ a, \(25-x^2=5^2-x^2=\left(5-x\right)\left(5+x\right)\)

b/ \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)

c/ \(9x^2+6xy+y^2=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

Xét \(\Delta OAE\) vuông tại E ta có :

\(CE^2=OC^2-OA^2\) (Định lí Py ta go) \(\left(1\right)\)

Xét \(\Delta OBF\) vuông tại F có :

\(BF^2=OB^2-OF^2\) (Đính lí Py ta go) \(\left(2\right)\)

Xét \(\Delta OAD\) vuông tại D có :

\(AD^2=OA^2-OD^2\) (Đính lí Py ta go) \(\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow AD^2+BF^2+CE^2=OC^2-OA^2+OB^2-ÒF^2+OA^2-OD^2\)

\(\Leftrightarrow AE^2+BF^2+CD^2=\left(AO^2-ÒD^2\right)+\left(OC^2-OF^2\right)+\left(OB^2-OD^2\right)\)

\(\Leftrightarrow AD^2+BF^2+CE^2=AE^2+CF^2+BD^2\left(đpcm\right)\)

\(8^7-2^{18}=\left(2^3\right)^7-2^{18}=2^{21}-2^{18}=2^{18}\left(2^3-1\right)=2^{17}.2.7=2^{17}.14⋮14\left(đpcm\right)\)

1/ Ta có :

\(\left|2x-\dfrac{1}{2}\right|\ge0\)

\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|-2017\ge-2017\)

\(\Leftrightarrow A\ge-2017\)

Dấu "=" xảy ra khi :

\(\left|2x-\dfrac{1}{2}\right|=0\)

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

Vậy ....

2/ Ta có :

\(\left|x+2017\right|\ge0\)

\(\Leftrightarrow-\left|x+2017\right|\le0\)

\(\Leftrightarrow2017-\left|x+2017\right|\le2017\)

\(\Leftrightarrow B\le2017\)

Dấu "=" xảy ra khi :

\(\left|x+2017\right|=0\Leftrightarrow x=-2017\)

Vậy..

a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

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

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

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

Vậy ..

c/ \(\left(2x-3\right)^3=-8\)

\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-3=-2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

d/ \(\left(3x-2\right)^5=-243\)

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

\(\Leftrightarrow3x-2=-3\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Vậy ...

e/ \(\left(x-1\right)^3=-125\)

\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x-1=-5\)

\(\Leftrightarrow x=-4\)

Vậy..

f/ \(\left(2x-1\right)^4=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

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

Vậy...

g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)

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

Vậy..

2/ Ta có :

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

Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)

3/ \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)

Mà \(x+y=7;xy=-3\)

\(\Leftrightarrow x^2+y^2=7^2-2.\left(-3\right)=49+6=55\)

a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)

Với mọi x ta có :

\(\left(x-3\right)^2\ge0\)

\(\Leftrightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-6x+10>0\)

b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)

Với mọi x ta có :

\(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)

\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)

c/ \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Với mọi x ta có :

\(\left(x+\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)

d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)

Với mọi x,y ta có :

\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)

\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)

Ta có :

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(\Leftrightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)

\(\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abc-acy-bcx-abz-acy-bcx}{a^2+b^2+c^2}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{bz-cy}{a}=0\\\dfrac{cx-az}{b}=0\\\dfrac{ay-bx}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)

a/ \(\left|x-3\right|=x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=x+1\\x-3=-x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-x=1+3\\x+x=-1+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=4\left(loại\right)\\2x=2\end{matrix}\right.\) \(\Leftrightarrow x=1\)

Vậy ...

b/ \(\left|x-2\right|=2x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2x+3\\x-2=-2x-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-x=-2-3\\x+2x=-3+2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy ,..

a/ Với mọi x, y ta có :

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\y^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|x-2\right|+y^2\ge0\)

\(\Leftrightarrow\left|x-2\right|+y^2+5\ge5\)

\(\Leftrightarrow A\ge5\)

Dấu "=" xảy ra khi :

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

Vậy...

b/ Với mọi x ta có :

\(\left|x-2\right|\ge0\)

\(\Leftrightarrow\left|x-2\right|-1\ge-1\)

\(\Leftrightarrow B\ge-1\)

Dấu "=" xảy ra khi :

\(\left|x-2\right|=0\Leftrightarrow x=2\)

Vậy ....

c/ Với mọi x ta có :

\(\left|1-x\right|\ge0\)

\(\Leftrightarrow2\left|1-x\right|\ge0\)

\(\Leftrightarrow2\left|1-x\right|+1\ge1\)

\(\Leftrightarrow C\ge1\)

Dấu "=" xảy ra khi :

\(\left|1-x\right|=0\Leftrightarrow x=1\)

Vậy ...

\(A=16x^2-24x+30\)

\(\Leftrightarrow A=\left(4x\right)^2-2.4x.3+3^2+21\)

\(\Leftrightarrow A=\left(4x-3\right)^2+21\)

Với mọi x ta có :

\(\left(4x-3\right)^2\ge0\)

\(\Leftrightarrow\left(4x+3\right)^2+21\ge21\)

\(\Leftrightarrow A\ge21\)

Dấu "=" xảy ra khi :

\(\left(4x+3\right)^2=0\Leftrightarrow x=-\dfrac{3}{4}\)

Vậy ...

\(A=1+3+3^2+....+3^{20}\)

\(\Leftrightarrow3A=3+3^2+...+3^{21}\)

\(\Leftrightarrow3A-A=\left(3+3^2+...+3^{21}\right)-\left(1+3+....+3^{20}\right)\)

\(\Leftrightarrow2A=3^{21}-1\)

\(\Leftrightarrow A=\dfrac{3^{21}-1}{2}\)

Mà \(B=3^{21}-1\)

\(\Leftrightarrow A< B\)

a/ \(10^{n+1}+6.10^n=10^n.10+6.10^n=10^n\left(10+6\right)=10^n.16\)

b/ \(90.10^n-10^{n+2}+10^{n+1}=90.10^n-10^n.10^2+10^n.10=10^n\left(90-100+10\right)=0\)

c/ \(2,5.5^{n-1}.10+5^n-6.5^{n-1}=2,5.5^n.\dfrac{1}{5}+5^n-6.5^n.\dfrac{1}{5}=5^n\left(\dfrac{1}{2}+1+\dfrac{6}{5}\right)=5^n.\dfrac{3}{2}\)

Ta có :

\(2x^2+4x+15=2\left(x^2+2x+\dfrac{15}{2}\right)=2\left(x^2+2.x.1+1^2+\dfrac{13}{2}\right)=2\left[\left(x+1\right)^2+\dfrac{13}{2}\right]=2\left(x+1\right)^2+13\)

Với mọi x ta có :

\(2\left(x+1\right)^2\ge0\)

\(\Leftrightarrow2\left(x+1\right)^2+13\ge13\)

Dấu "=" xảy ra khi :

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

Vậy ...

\(A=2+2^2+2^3+....+2^{99}\)

\(\Leftrightarrow2A=2^2+2^3+...+2^{99}+2^{100}\)

\(\Leftrightarrow2A-A=\left(2^2+2^3+....+2^{100}\right)-\left(2+2^2+....+2^{99}\right)\)

\(\Leftrightarrow A=2^{100}-2\)

Mà \(B=2^{100}-2\)

\(\Leftrightarrow A=B\)

Đặt :

\(\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=dk\\b=ek\\c=fk\end{matrix}\right.\)

\(\Leftrightarrow P=\dfrac{ax+bx+c}{dx^2+ẽx+f}=\dfrac{dkx^2+ekx+fk}{dx^2+ex+f}=\dfrac{k\left(dx^2+ex+f\right)}{dx^2+ex+f}=k\)

Vậy nếu \(\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}\) thì P k phụ thuộc vào x

\(\left(x-3\right)^2-4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{4}\\x-3=-\sqrt{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Vậy ..

có nhầm đề k cậu?

\(\left|x+1\right|=2x-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=0\left(loại\right)\end{matrix}\right.\)

Vậy ....

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy ...

\(2^{x+1}.3^y=12^x\)

\(\Leftrightarrow2^{x+1}.3^y=3^x.2^{2x}\)

\(\Leftrightarrow x+1=2x,y=x\)

\(\Leftrightarrow x=y=1\)

Vậy..

câu b mình có sửa lại đề :)

Buồn nhiều :)