### Toán Lớp 9: √x-5=7 CM:√11+6√2-√2=3 Tính:√144.√49/64.√0,01 PTTNT:7+4√3

Toán Lớp 9: √x-5=7
CM:√11+6√2-√2=3
Tính:√144.√49/64.√0,01
PTTNT:7+4√3

TRẢ LỜI

1. Giải đáp:
a) ĐKXĐ: $x\geq5$
Ta có: $\sqrt{x-5}=7$ ⇔ $x-5=49$ ⇔ $x=54$ ( thỏa mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là S = { 54 }
b) Ta có: $\sqrt{11+6\sqrt{2}}-\sqrt{2}$ $=$ $\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{2}$
$=$ $\sqrt{(3+\sqrt{2})^2}-\sqrt{2}$ $=$ $|3+\sqrt{2}|-\sqrt{2}$
$=$ $3+\sqrt{2}-\sqrt{2}$ $=$ $3$ ( đpcm )
c) Ta có: $\sqrt{144}$ $.\sqrt{\frac{49}{64}}$ $.\sqrt{0,01}$
$=$ $\sqrt{12^2}$ $.\sqrt{(\frac{7}{8})^2}$ $.\sqrt{(0,1)^2}$
$=$ $|12|.|\frac{7}{8}|.|0,1|$ $=$ $12.\frac{7}{8}.0,1$ $=$ $\frac{21}{20}$
d) Ta có: $7+4\sqrt{3}=4+2.2\sqrt{3}+3=(2+\sqrt{3})^2$

Trả lời
2. bichquyenlien
$\begin{array}{l} \sqrt x – 5 = 7\left( {x \ge 0} \right)\\ \Leftrightarrow \sqrt x = 12\\ \Leftrightarrow x = 144\\ \Rightarrow S = \left\{ {144} \right\}\\ \sqrt {11 + 6\sqrt 2 } – \sqrt 2 \\ = \sqrt {9 + 2 + 2.3.\sqrt 2 } – \sqrt 2 \\ = \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} – \sqrt 2 \\ = 3 + \sqrt 2 – \sqrt 2 \\ = 3\\ \sqrt {144} .\sqrt {\dfrac{{49}}{{64}}} .\sqrt {0,01} \\ = \sqrt {{{12}^2}} .\sqrt {{{\left( {\dfrac{7}{8}} \right)}^2}} .\sqrt {{{\left( {0,1} \right)}^2}} \\ = 12.\dfrac{7}{8}.0,1\\ = \dfrac{{21}}{{20}}\\ 7 + 4\sqrt 3 = 4 + 3 + 2.2.\sqrt 3 = {\left( {2 + \sqrt 3 } \right)^2} \end{array}$

Trả lời