\(B=\dfrac{2^2}{x}+\dfrac{3^2}{y}\ge\dfrac{\left(2+3\right)^2}{x+y}=25\)

\(B_{min}=25\) khi \(\left(x;y\right)=\left(\dfrac{2}{5};\dfrac{3}{5}\right)\)

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6

\(\frac{x}{-4}=-\frac{9}{x}\)

\(\Rightarrow x^2=36\)

\(\Rightarrow x=\pm6\)

Vậy x= 6 ; x = - 6

Vì \(\frac{x}{-4}\)=\(\frac{-9}{x}\)

\(\Rightarrow\)x.x=-4.-9

\(\Rightarrow\)\(x^2\)=36

\(\Rightarrow\)\(\left[\begin{array}{nghiempt}x=-6\\x=6\end{array}\right.\)

Vậy x=-6 hoặc x=6

Vì c/-4=-9/x =>x^2=36 =>x=6 hoặc x=-6 Vậy x=6 hoặc x=-6

Sửa đề: Tìm các giá trị nguyên X thỏa mãn

3: \(\frac{-5}{11}<\frac{9}{x}<\frac{-5}{12}\)

=>\(\frac{-45}{99}<\frac{-45}{-5x}<\frac{-45}{108}\)

=>\(\frac{45}{99}>\frac{45}{-5x}>\frac{45}{108}\)

=>99<-5x<108

=>\(-\frac{99}{5}>x>-\frac{108}{5}\)

mà x nguyên

nên x∈{-20;-21}

4: Ta có: \(\frac{-11}{13}<\frac{9}{x}<\frac{-11}{15}\)

=>\(\frac{-99}{117}<\frac{-99}{-11x}<\frac{-99}{135}\)

=>\(\frac{99}{117}>\frac{99}{-11x}>\frac{99}{135}\)

=>117<-11x<135

=>\(-\frac{117}{11}>x>-\frac{135}{11}\)

mà x nguyên

nên x∈{-11;-12}

5: Ta có: \(\frac{-4}{5}<\frac{9}{x}<\frac{-4}{7}\)

=>\(\frac{-36}{45}<\frac{-36}{-4x}<\frac{-36}{28}\)

=>\(\frac{36}{45}>\frac{36}{-4x}>\frac{36}{28}\)

=>45<-4x<28

=>\(-\frac{45}{4}>x>-\frac{28}{4}\)

mà x nguyên

nên x∈∅

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

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)

Lời giải:
$(x+\frac{4}{9})^2\geq 0$ (do bình phương 1 số thì không âm)

$\frac{-49}{144}< 0$

Do đó: $(x+\frac{4}{9})^2> \frac{-49}{144}$ với mọi $x$ nên pt trên vô nghiệm.

Ta có: \(\left(x+\dfrac{4}{9}\right)^2=-\dfrac{49}{144}\)

mà \(\left(x+\dfrac{4}{9}\right)^2\ge0\forall x\)

nên \(x\in\varnothing\)