Bài 5:

a) Để hpt có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{2}{m}\Leftrightarrow m\ne\pm2\) 

\(\left\{{}\begin{matrix}mx+2y=m+1\\2x+my=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\2x+m\cdot\dfrac{m-mx+1}{2}=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\2x+\dfrac{m^2-m^2x+m}{2}=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\4x+m^2-m^2x+m=4m-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\\left(m^2-4\right)x=m^2-3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-m\cdot\dfrac{m-1}{m+2}+1}{2}=\dfrac{\dfrac{m\left(m+2\right)-m\left(m-1\right)+m+2}{m+2}}{2}=\dfrac{2m+1}{m+2}\\x=\dfrac{m^2-3m+2}{m^2-4}=\dfrac{m-1}{m+2}\end{matrix}\right.\) 

Để x,y nguyên thì \(\dfrac{m-1}{m+2};\dfrac{2m+1}{m+2}\) phải nguyên 

+) Ta có: \(\dfrac{m-1}{m+2}=\dfrac{m+2-3}{m+2}=1-\dfrac{3}{m+2}\)

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (1)

+) Ta có: \(\dfrac{2m+1}{m+2}=\dfrac{2m+4-3}{m+2}=2-\dfrac{3}{m+2}\)

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (2) 

Từ (1) và (2) => m ∈ {1; -1; 3; -3} 

Bài 4 

a, \(\left\{{}\begin{matrix}-2\sqrt{3}x+3\sqrt{5}y=-21\\4x-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21-3\sqrt{5}y}{-2\sqrt{3}}\\\dfrac{4\left(21-3\sqrt{5}y\right)}{-2\sqrt{3}}-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow84-21\sqrt{5}y+12y=-12\left(2+\sqrt{5}\right)\)

\(\Leftrightarrow84+y\left(-21\sqrt{5}+12\right)=-24-12\sqrt{5}\Leftrightarrow y=\dfrac{-108-12\sqrt{5}}{-21\sqrt{5}+12}\)

\(\Rightarrow x=\dfrac{\dfrac{\left(21-3\sqrt{5}\right).\left(-108-12\sqrt{5}\right)}{-21\sqrt{5}+12}}{-2\sqrt{3}}\)

b, \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-2\right)^2=\left(x+1\right)^2+1+\left(y+1\right)^2\\\left(x-y-3\right)^2=\left(x-y-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2-\left(x+1\right)^2=1+\left(y+1\right)^2-\left(y-2\right)^2\\\left(x-y-3-x+y+1\right)\left(x-y-3+x-y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-4x-1=-\left(2y-1\right)\\-2\left(2x-2y-4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+2y=2\\x-y-2=0\end{matrix}\right.\)

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

Vậy hpt vô nghiệm 

Diện tích sắt cần để làm lồng sắt là:

\(2\times\left(4+1\right)\times2+2\times4\times1=28\left(m^2\right)\)

Làm lồng sắt hết số tiền là:

\(28\times39000=1092000\left(đ\right)\)

ĐS: ...

e

Bài 1 

\(1,\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3=x^3+3x^2+3x+1-x^2+16-x^3=2x^2+3x+17\)

2, \(\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8=x^3+6x^2+12x+8-x\left(x^2-9\right)-12x^2-8\)

\(=x^3+6x^2+12x+8-x^3+9x-12x^2-8=-6x^2+21x\)

3, \(\left(x-2\right)^3-x\left(x-2\right)\left(x+2\right)+6x\left(x-3\right)\)

\(=x^3-6x^2+12x-8-x\left(x^2-4\right)+6x^2-18x\)

\(=x^3+12x-8-x^3+4x-18x=2x=8\)

4, \(x\left(x-5\right)\left(x+5\right)-\left(x-5\right)^3+100x\)

\(=\left(x-5\right)\left[x^2+5x-\left(x-5\right)^2\right]+100x\)

\(=\left(x-5\right)\left(x^2+5x-x^2+10x-25\right)+100x=\left(x-5\right)\left(15x-25\right)+100x\)

\(=15x^2-100x+125+100x=15x^2+125\)

5, \(\left(x-3y\right)^3-\left(x-2y\right)\left(2y+x\right)=x^3-9x^2y+27xy^2-27y^3-x^2+4y^2\)

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

\(=\left(x+2y\right)\left[-\left(x+2y\right)^2+2xy-x^2\right]=\left(x+2y\right)\left(-x^2-4xy-4y^2+2xy-x^2\right)=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)\)

\(=\left(x+2y\right)\left(-2x^2-2xy-4y^2\right)=-2x^3-2x^2y-4xy^2-4x^2y+4xy^2-8y^3=-2x^3-6x^2y-8y^3\)

7, \(-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)

\(=-\left(8x^3-12x^2y+6xy^2-y^3\right)-x\left(4x^2-4xy+y^2\right)-y^3=-12x^3+16x^2y-7xy^2\)

8, \(-x\left(x-y\right)^2+\left(x-y\right)^3+y^2\left(y-2x\right)\)

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

Bài 6:

1)

\(8x^3-12x^2+6x-1=0\\ \Leftrightarrow\left(2x-1\right)^3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

2) 

\(x^3-6x^2+12x-8=0\\ \Leftrightarrow\left(x-2\right)^3=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)

3) 

\(x^2-8x+16=5\left(4-x\right)^3\\ \Leftrightarrow\left(x-4\right)^2-5\left(4-x\right)^3=0\\ \Leftrightarrow\left(4-x\right)^2-5\left(4-x\right)^3=0\\ \Leftrightarrow\left(4-x\right)^2\left[1-5\left(4-x\right)\right]=0\\ \Leftrightarrow\left(4-x\right)^2\left(5x-19\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{19}{5}\end{matrix}\right.\)

4) 

\(\left(2-x\right)^3=6x\left(x-2\right)\\ \Leftrightarrow\left(2-x\right)^3-6x\left(x-2\right)=0\\ \Leftrightarrow\left(2-x\right)^3+6x\left(2-x\right)=0\\ \Leftrightarrow\left(2-x\right)\left[\left(2-x\right)^2+6x\right]=0\\ \Leftrightarrow\left(2-x\right)\left(4-4x+x^2+6x\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow2-x=0\\ \Leftrightarrow x=2\)

(vì x^2+2x+4=x^2+2x+1+3=(x+1)^2+3>0) 

Để số 2014xy chia hết cho 2 và 5 nên y tận cùng là 0

ta có : 2+0+1+4+x+0 = x +7 

vậy để 2014xy chia hết cho 9 thì x+7 chia hết cho 9

vậy x = 2 

vậy số cần tìm là 201 420

Đề là chia hết cho 9 chứ không phải 3 mà bạn.

Thiếu dữ kiện nha. Ko có kết quả thì làm sao mà tính đc?

Bạn kiểm tra lại đề nhé

Đặt: 

\(A=\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{25^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{\left(5^2\right)^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\\ 5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{19}}\\ 5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{19}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\right)\\4A =1-\dfrac{1}{5^{20}}\\ 4A=\dfrac{5^{20}-1}{5^{20}}\\ A=\dfrac{5^{20}-1}{4\cdot5^{20}}\)

ê

\(\left[a+\left(b+c\right)\right]^2\\=a^2+2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2+2ab+2ac+\left(b^2+2bc+c^2\right)\\ =a^2+b^2+c^2+2ab+2ac+2bc\)

ê

\(\left(x+\dfrac{1}{3}\right)+\left(x+\dfrac{1}{9}\right)+\left(x+\dfrac{1}{27}\right)+\left(x+\dfrac{1}{81}\right)=\dfrac{56}{81}\\ x+\dfrac{1}{3}+x+\dfrac{1}{9}+x+\dfrac{1}{27}+x+\dfrac{1}{81}=\dfrac{56}{81}\\ \left(x+x+x+x\right)+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\right)\\ x\times\left(1+1+1+1\right)+\left(\dfrac{27}{81}+\dfrac{9}{81}+\dfrac{3}{81}+\dfrac{1}{81}\right)=\dfrac{56}{81}\\ x\times4+\dfrac{40}{81}=\dfrac{56}{81}\\ x\times4=\dfrac{56}{81}-\dfrac{40}{81}\\ x\times4=\dfrac{16}{81}\\x=\dfrac{16}{81}:4\\ x=\dfrac{4}{81}\)

ê

Đặt P ở trọng tâm tam giác nha. vì trọng tâm cách đều 3 đỉnh nên như thế là tiết kiệm nhất nhé

ê

\(\dfrac{7}{8}-\dfrac{7}{16}-\dfrac{11}{32}\)

\(=\dfrac{14}{16}-\dfrac{7}{16}-\dfrac{11}{32}\)

\(=\dfrac{7}{16}-\dfrac{11}{32}\)

\(=\dfrac{14}{32}-\dfrac{11}{32}\)

\(=\dfrac{3}{32}\)

\(#NqHahh\)

`7/8-7/16-11/32`

`=28/32-14/32-11/32`

`=3/32`