Áp dụng BĐT Bunhiacopxki (cho tất cả các bài):

1.

\(\left(3x+4y\right)^2\le\left(3^2+4^2\right)\left(x^2+y^2\right)=25\)

\(\Rightarrow\left|3x+4y\right|\le5\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{3}{5};\dfrac{4}{5}\right)\)

2.

\(\left(x+2y\right)^2=\left(1.x+\sqrt{2}.\sqrt{2y}\right)^2\le\left(1+2\right)\left(x^2+2y^2\right)=3\)

\(\Rightarrow\left|x+2y\right|\le\sqrt{3}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{\sqrt{3}};\dfrac{1}{\sqrt{3}}\right)\)

4.

a.

Áp dụng Bunhiacopxki:

\(\left(b+c+c+a+a+b\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow2\left(a+b+c\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

\(ZnO,BaO,K_2O\)

\(ZnCl_2,BaCl_2,KCl\)

\(Zn\left(OH\right)_2,Ba\left(OH\right)_2,KOH\)

\(ZnSO_4,BaSO_4,K_2SO_4\)

\(Zn\left(NO_3\right)_2,Ba\left(NO_3\right)_2,KNO_3\)

\(ZnCO_3,BaCO_3,K_2CO_3\)

\(Zn_3\left(PO_4\right)_2,Ba_3\left(PO_4\right)_2,K_3PO_4\)

Còn câu b

\(\Leftrightarrow3^{x-3}=27\)

=>x-3=3

hay x=6

mail mình là:

hoangclgt2003@gmail.com

quynhphamtrinh@gmail.com

ĐKXĐ: x<>0

Ta có: \(8x^2+\frac{2}{x^2}\)

\(=2\left(4x^2+\frac{1}{x^2}\right)\)

\(=2\left\lbrack4x^2+\frac{1}{x^2}-2\cdot2x\cdot\frac{1}{x}+2\cdot2x\cdot\frac{1}{x}\right\rbrack=2\left\lbrack\left(2x-\frac{1}{x}\right)^2+4\right\rbrack\)

Ta có: \(8x^2+\frac{2}{x^2}-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left\lbrack\left(\left|2x-\frac{1}{x}\right|^{}\right)^2+4\right\rbrack-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2+8-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2-9\cdot\left|2x-\frac{1}{x}\right|+7=0\)

=>\(\left(2\cdot\left|2x-\frac{1}{x}\right|-7\right)\left(\left|2x-\frac{1}{x}\right|-1\right)=0\)

TH1: \(2\cdot\left|2x-\frac{1}{x}\right|-7=0\)

=>\(\left|2x-\frac{1}{x}\right|=\frac72\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=\frac72\\ 2x-\frac{1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=\frac72\\ \frac{2x^2-1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-2=7x\\ 4x^2-2=-7x\end{array}\right.\)

=>\(\left[\begin{array}{l}4x^2-7x-2=0\\ 4x^2+7x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-8x+x-2=0\\ 4x^2+8x-x-2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}\left(x-2\right)\left(4x+1\right)=0\\ \left(x+2\right)\left(4x-1\right)=0\end{array}\right.\)

=>x∈{2;-2;-1/4;1/4}

TH2: \(\left|2x-\frac{1}{x}\right|-1=0\)

=>\(\left|2x-\frac{1}{x}\right|=1\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=1\\ 2x-\frac{1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=1\\ \frac{2x^2-1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x^2-1-x=0\\ 2x^2+x-1=0\end{array}\right.\)

=>\(\left[\begin{array}{l}2x^2-2x+x-1=0\\ 2x^2+2x-x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(x-1\right)\left(2x+1\right)=0\\ \left(x+1\right)\left(2x-1\right)=0\end{array}\right.\)

=>x∈{1;-1;-1/2;1/2}

ĐKXĐ: x<>0

Ta có: \(8x^2+\frac{2}{x^2}\)

\(=2\left(4x^2+\frac{1}{x^2}\right)\)

\(=2\left\lbrack4x^2+\frac{1}{x^2}-2\cdot2x\cdot\frac{1}{x}+2\cdot2x\cdot\frac{1}{x}\right\rbrack=2\left\lbrack\left(2x-\frac{1}{x}\right)^2+4\right\rbrack\)

Ta có: \(8x^2+\frac{2}{x^2}-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left\lbrack\left(\left|2x-\frac{1}{x}\right|^{}\right)^2+4\right\rbrack-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2+8-9\cdot\left|2x-\frac{1}{x}\right|-1=0\)

=>\(2\cdot\left(\left|2x-\frac{1}{x}\right|\right)^2-9\cdot\left|2x-\frac{1}{x}\right|+7=0\)

=>\(\left(2\cdot\left|2x-\frac{1}{x}\right|-7\right)\left(\left|2x-\frac{1}{x}\right|-1\right)=0\)

TH1: \(2\cdot\left|2x-\frac{1}{x}\right|-7=0\)

=>\(\left|2x-\frac{1}{x}\right|=\frac72\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=\frac72\\ 2x-\frac{1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=\frac72\\ \frac{2x^2-1}{x}=-\frac72\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-2=7x\\ 4x^2-2=-7x\end{array}\right.\)

=>\(\left[\begin{array}{l}4x^2-7x-2=0\\ 4x^2+7x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}4x^2-8x+x-2=0\\ 4x^2+8x-x-2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}\left(x-2\right)\left(4x+1\right)=0\\ \left(x+2\right)\left(4x-1\right)=0\end{array}\right.\)

=>x∈{2;-2;-1/4;1/4}

TH2: \(\left|2x-\frac{1}{x}\right|-1=0\)

=>\(\left|2x-\frac{1}{x}\right|=1\)

=>\(\left[\begin{array}{l}2x-\frac{1}{x}=1\\ 2x-\frac{1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{2x^2-1}{x}=1\\ \frac{2x^2-1}{x}=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x^2-1-x=0\\ 2x^2+x-1=0\end{array}\right.\)

=>\(\left[\begin{array}{l}2x^2-2x+x-1=0\\ 2x^2+2x-x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(x-1\right)\left(2x+1\right)=0\\ \left(x+1\right)\left(2x-1\right)=0\end{array}\right.\)

=>x∈{1;-1;-1/2;1/2}

Câu bữa trc mình gửi á

Câu 6)

\(U_{13}=U_{12}+U_{23}=2,4+2,5=4,9V\\ U_{23}=U_{13}-U_{12}=5,4V\\ U_{12}=U_{13}-U_{23}=11,7V\) 

Câu 7)

\(I_1=I_2=0,25A\\ I_3=I-I_1\left(I_2\right)=0,1A\\ U=U_1=U_3=9V\\ U_2=U-U_1=3V\)

thnk nha

 1/x + 1/y = 1/2 
<=> 2/x + 2/y =1 
<=> 2x + 2y = xy 
<=> xy - 2x - 2y+4 =4 
<=> (x-2).(y-2) =4 
ta có các cặp (2;2),(-2;-2),(1;4), (-1;-4) 
sau đó thử từng cặp một bạn nhé

cảm ơn bạn Thắng Hoàng nhưng có cái này mình ko hiểu lắm mấu chứa ẩn thì làm sao mà khử dc