\(b,=\left|\dfrac{17}{6}-\dfrac{35}{6}\right|+1=3+1=4\\ c,=\dfrac{7^3\left(7^2-4\right)}{45}=\dfrac{7^3\cdot45}{45}=7^3=343\)

a: \(7+\left(\dfrac{7}{12}-\dfrac{1}{2}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+\dfrac{7}{12}-\dfrac{1}{12}-\dfrac{1}{2}+3-5\)

\(=7+1-2\)

=6

c) \(1-\left\{1:\left[2^3+1-\left(-\dfrac{1}{2}\right)^2\right]\right\}\)\(=1-\left[1:\left(8+1-\dfrac{1}{4}\right)\right]=1-\left(1:\dfrac{35}{4}\right)=1-\dfrac{4}{35}\)\(=\dfrac{35-4}{35}=\dfrac{31}{35}\)

Với `x >= 0,x \ne 1` có:

`C=A/B=A:B=[\sqrt{x}+1]/[x+\sqrt{x}+1]:(\sqrt{x}/[x\sqrt{x}-1]+1/[\sqrt{x}-1])`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1]:[\sqrt{x}+x+\sqrt{x}+1]/[(\sqrt{x}-1)(x+\sqrt{x}+1)]`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[x+2\sqrt{x}+1]`

`C=[\sqrt{x}+1]/[x+\sqrt{x}+1].[(\sqrt{x}-1)(x+\sqrt{x}+1)]/[(\sqrt{x}+1)^2]`

`C=[\sqrt{x}-1]/[\sqrt{x}+1]`

1.Thế \(x=4\) vào A, ta được:

\(A=\dfrac{\sqrt{4}+1}{4+\sqrt{4}+1}=\dfrac{2+1}{4+2+1}=\dfrac{3}{7}\)

2.

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

\(B=\dfrac{\sqrt{x}}{\sqrt{x}^3-1}+\dfrac{1}{\sqrt{x}-1}\)

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

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

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

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

\(C=\dfrac{A}{B}\)

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

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

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

\(C=A:B=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+x+\sqrt{x}+1}{x\sqrt{x}-1}\right)\)

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

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

a: Ta có: \(\widehat{xBC}=\widehat{ABM}\)

mà \(\widehat{ABM}=\widehat{NMB}\)

nên \(\widehat{xBC}=\widehat{BMN}\)

c) What does your sister do in the morning ?

e) What did he do last Sunday ?

c) What does your sister do in the morning ?

e) What did he do last Sunday ?

ngu