y x(5+7)=120

y x 12=120

y=10

X x [5+7] =120

X x 12=120

X =10

Bài 2:

a: \(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\cdot\sqrt{x}+x-\sqrt{x}-1}\)

\(=\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)

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

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

Ta có: \(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\)

\(=\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)

Ta có: \(C=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\cdot\sqrt{x}+x-\sqrt{x}-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)

\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}:\frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

a.

ĐKXĐ: \(x\ge2\)

\(\dfrac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}+\dfrac{\sqrt{x-1}-\sqrt{x-2}}{\left(\sqrt{x-1}+\sqrt{x-2}\right)\left(\sqrt{x-1}-\sqrt{x-2}\right)}=1\)

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

\(\Leftrightarrow\sqrt{x}-\sqrt{x-2}=1\)

\(\Leftrightarrow\sqrt{x}=1+\sqrt{x-2}\)

\(\Leftrightarrow x=1+x-2+2\sqrt{x-2}\)

\(\Leftrightarrow2\sqrt{x-2}=1\)

\(\Leftrightarrow x-2=\dfrac{1}{4}\)

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

b

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\dfrac{x-1}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}=\dfrac{x-1}{2}\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|=\dfrac{x-1}{2}\)

Đặt \(\sqrt{x-1}=t\ge0\Rightarrow\left|t-1\right|+\left|t+1\right|=\dfrac{t^2}{2}\)

TH1: \(0\le t\le1\) pt trở thành:

\(1-t+t+1=\dfrac{t^2}{2}\Rightarrow t^2=4\)

\(\Rightarrow\left[{}\begin{matrix}t=2>1\left(ktm\right)\\t=-2< 0\left(ktm\right)\end{matrix}\right.\)

TH2: \(t>1\) pt trở thành:

\(t-1+t+1=\dfrac{t^2}{2}\Rightarrow t^2=2t\Rightarrow\left[{}\begin{matrix}t=0< 1\left(ktm\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-1}=2\Rightarrow x=5\)

137/10ha = ….m²

Đáp án 137 000 m²

HT

137/10ha = 137/100000 m²

k chắc đâu nhé:))

A

A

A

A

Ta có: \(x\cdot62+x\cdot21=1909\)

\(\Leftrightarrow x\cdot83=1909\)

hay x=23

62x + 21x = 1909

<=> 83x = 1909

<=> x = \(\dfrac{1909}{83}\) = 23

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

\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{10}}\)

đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=2-\frac{1}{2^{10}}\)