\(e,=\dfrac{\left(3+\sqrt{2}\right)\left(2\sqrt{2}+1\right)}{7}-\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}\\ =\dfrac{7\sqrt{2}+7}{7}-\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1-\sqrt{2}-1=0\)

\(f,=\sqrt{\dfrac{\left(2\sqrt{3}-3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}}\left(2+\sqrt{3}\right)\\ =\dfrac{\left(2\sqrt{3}-3\right)\left(2+\sqrt{3}\right)}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}=1\)

\(h,=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{20-9}}\left(\sqrt{2}+\sqrt{10}\right)\\ =\sqrt{\dfrac{2\left(33-11\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{\dfrac{22\left(3-\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}+1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)

Lời giải:

b. Tam giác $ABC$ vuông tại $A$ và $C=45^0$ nên:

 $B=90^0-C=90^0-45^0=45^0$

Do đó, tam giác $ABC$ vuông cân tại $A$

$\Rightarrow AC=AB=50$ (cm)

Áp dụng định lý Pitago: $BC=\sqrt{AB^2+AC^2}=\sqrt{50^2+50^2}=50\sqrt{2}$ (cm)

f.

Theo định lý Pitago: $AC=\sqrt{BC^2-AB^2}=\sqrt{7^2-5^2}=2\sqrt{6}$ (cm)

$\sin B=\frac{AC}{BC}=\frac{2\sqrt{6}}{7}$

$\Rightarrow B=44,42^0$

$C=90^0-B=90^0-44,42^0=45,58^0$

b) Xét ΔABC vuông tại A có \(\widehat{C}=45^0\)(gt)

nên ΔABC vuông cân tại A(Định nghĩa tam giác vuông cân)

Suy ra: \(\widehat{B}=45^0\) và AC=50(cm)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=50^2+50^2=5000\)

hay \(BC=50\sqrt{2}\left(cm\right)\)

a) Ta có: ΔABC vuông tại A(gt)

nên \(\widehat{B}+\widehat{C}=90^0\)

hay \(\widehat{B}=60^0\)

Xét ΔABC vuông tại A có 

\(AB=AC\cdot\tan30^0\)

\(=100\cdot\dfrac{\sqrt{3}}{3}\)

\(=\dfrac{100\sqrt{3}}{3}\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=100^2+\left(\dfrac{100\sqrt{3}}{3}\right)^2=\dfrac{40000}{3}\)

hay \(AC=\dfrac{200\sqrt{3}}{3}\left(cm\right)\)

a) Ta có: \(x-\dfrac{1}{2}=\left|\dfrac{3}{7}\right|\)

nên \(x-\dfrac{1}{2}=\dfrac{3}{7}\)

hay \(x=\dfrac{3}{7}+\dfrac{1}{2}=\dfrac{6}{14}+\dfrac{7}{14}=\dfrac{13}{14}\)

b) Ta có: |x-1|=0

nên x-1=0

hay x=1

c) Ta có: \(\left|x+1\right|\ge0\forall x\)

\(\left|y-2\right|\ge0\forall y\)

Do đó: \(\left|x+1\right|+\left|y-2\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi x=-1 và y=2

d) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}\)

mà x-y=-4

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{-4}{-2}=2\)

Do đó: x=6; y=10

e) Ta có: 3x=4y

nên \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}\)

Đặt \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k\\y=\dfrac{1}{4}k\end{matrix}\right.\)

Ta có: xy=48

nên \(\dfrac{1}{3}k\cdot\dfrac{1}{4}k=48\)

\(\Leftrightarrow k^2\cdot\dfrac{1}{12}=48\)

\(\Leftrightarrow k^2=48\cdot12=576\)

hay \(k\in\left\{24;-24\right\}\)

Trường hợp 1: k=24

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k=\dfrac{1}{3}\cdot24=8\\y=\dfrac{1}{4}k=\dfrac{1}{4}\cdot24=6\end{matrix}\right.\)

Trường hợp 2: k=-24

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}k=\dfrac{1}{3}\cdot\left(-24\right)=-8\\y=\dfrac{1}{4}k=\dfrac{1}{4}\cdot\left(-24\right)=-6\end{matrix}\right.\)

a, Ta có
     x- \(\dfrac{1}{2}\)= \(\left|\dfrac{3}{7}\right|\) 

⇒  x- \(\dfrac{1}{2}\)= \(\dfrac{3}{7}\)

⇒  x= \(\dfrac{13}{14}\)

   Vậy...

b, \(\left|x-1\right|\)= 0

    Ta có:

    \(\left|x-1\right|\)  \(\ge\) 0 với \(\forall\) x

    Dấu bằng xảy ra khi:

    x- 1= 0

⇒ x= 1

   Vậy....

c, \(\left|x+1\right|+\left|y-2\right|=0\)

    Ta có:

    \(\left|x+1\right|\ge0\) với ∀ x

    \(\left|y-2\right|\ge0\) với \(\forall\) y

\(\Rightarrow\)\(\left|x+1\right|\ge0\) + \(\left|y-2\right|\ge0\) ≥ 0 với \(\forall x,y\)

   Dấu bằng xảy ra khi:

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

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

⇒\(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

   Vậy.....

Đề bài đâu rồi em?

c: \(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}-1+\dfrac{x+4}{2016}-1=0\)

=>x-2012=0

hay x=2012

d: \(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

=>x-100=0

hay x=100

a: \(A=\dfrac{x}{x-4}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

b: Thay x=36 vào biểu thức \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\), ta có

\(\Rightarrow\dfrac{\sqrt{36}}{\sqrt{36}-2}\Rightarrow\dfrac{6}{6-2}\Rightarrow\dfrac{6}{4}\Rightarrow\dfrac{3}{2}\)

Tiếp của anh Nguyễn Lê Phước Thịnh:

b. Thay x = 36 vào A, ta được:

A = \(\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c. Ta có: 

\(\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\left(ĐK:x\ge0;x\ne2\right)\)

<=> \(-3\sqrt{x}=\sqrt{x}-2\)

<=> \(2=4\sqrt{x}\)

<=> \(\sqrt{x}=\dfrac{1}{2}\)

<=> \(x=\dfrac{1}{4}\left(TM\right)\)

e. Ta có: 

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

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

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

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

<=> \(\sqrt{x}=-2\sqrt{x}-2\)

<=> \(\sqrt{x}+2\sqrt{x}=-2\)

<=> \(3\sqrt{x}=-2\)

<=> \(\sqrt{x}=\dfrac{-2}{3}\)

<=> \(x=\dfrac{4}{9}\left(TM\right)\)

\(b,\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{\sqrt{15}}=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(d,\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\left(\sqrt{ab}-\sqrt{bc}\right)}=\sqrt{ab}+\sqrt{bc}=\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)\)

\(e,\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\left(\sqrt{\dfrac{a}{b}+\dfrac{2b.\sqrt{ab}}{b}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\sqrt{a}\sqrt{a+2b\sqrt{ab}}+b\sqrt{a^2}\)

\(=a\sqrt{a^2+2ab\sqrt{ab}}+ab\)

\(=a\left(\sqrt{a^2+2ab\sqrt{ab}}+b\right)\)

\(f,\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)

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

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

\(=\left(a-1\right)^2=a^2-2a+1\)