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a, \(=2\sqrt{7}-8+15\sqrt{7}-12=17\sqrt{7}-20\)
b, \(=2\sqrt{2}-10\sqrt{2}+4\sqrt{2}=-4\sqrt{2}\)
c, \(=\frac{3}{8}.\frac{4}{3}-2.\frac{2}{5}=\frac{1}{2}-\frac{4}{5}=-\frac{3}{10}\)
d, \(\sqrt{\left(\sqrt{3-1}\right)^2}-\sqrt{\left(\sqrt{3-2}\right)^2}=\sqrt{3-1}-\sqrt{3-2}=\sqrt{2}-\sqrt{1}=\sqrt{2}-1\)
e, \(\sqrt{2-3}\) không tồn tại
bài 2 rút gọn :
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
= \(\left|1-\sqrt{2}\right|+\left|\sqrt{2}-3\right|\)
=\(\sqrt{2}-1+3-\sqrt{2}\)
=2
b) \(\sqrt{4-2\sqrt{3}}+\sqrt{7}-\sqrt{48}\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{3}-1+\sqrt{7}-4\sqrt{3}\)
= \(\sqrt{7}-3\sqrt{3}+1\)
c)
\(a^3=16-8\sqrt{5}+16+8\sqrt{5}+3.\sqrt[3]{16^2-8^2.5}a\)
\(a^3=32+3.\sqrt[3]{4^3\left(4-5\right)}a=32-12a\)
\(f\left(x\right)=\left[\left(32-12a\right)+12a-31\right]^{2016}=1^{2016}=1\)
a=\(\sqrt[3]{16-8\sqrt{5}}\)+\(\sqrt[3]{16+8\sqrt{5}}\)
=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}+\sqrt[3]{1+3\sqrt{5}+15+5\sqrt{5}}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}+\sqrt[3]{\left(1+\sqrt{5}\right)^3}\)
=1-\(\sqrt{5}+1+\sqrt{5}\)=2
thay vào ta được f(a)=(8+24-31)2016=(-1)2016=1
Bài 1 :
a)\(\sqrt{-2\text{x}+3}\) <=> -2x+3 \(\ge\)0 <=> -2x \(\ge\) -3 <=> x\(\le\) \(\frac{3}{2}\)
b)\(\sqrt{\frac{4}{x+3}}< =>x+3>0< =>x>-3\)
Bài 2 :
a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)
b)\(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+\left|2-\sqrt{3}\right|=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
c) \(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
Bài 3 :
a) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
VT = \(\sqrt{5-2.2.\sqrt{5}+2^2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{5}\right)^2-4\sqrt{5}+2^2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
=|\(\sqrt{5-2}\)| -\(\sqrt{5}\)
= \(\sqrt{5}-2-\sqrt{5}\)
= -2 = VP
b)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
VT = \(\sqrt{7+2.4.\sqrt{7}+4^2}-\sqrt{7}\)
= \(\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}\)
= |\(\sqrt{7}+4\)| -\(\sqrt{7}\)
=\(\sqrt{7}+4-\sqrt{7}\)
= 4 =VP
c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
VT = \(16-8\sqrt{7}+7\)
= 23 - \(8\sqrt{7}\) = VP
Bài 4:
a)\(\frac{x^2-5}{x+\sqrt{5}}=\frac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\frac{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
Tương tự
Bài 5 :
a) \(\sqrt{x^2+6\text{x}+9}=3\text{x}-1\)
=> \(\sqrt{\left(x+3^2\right)}\) = 3x-1
=> x+3 = 3x-1
+) x+3 =3x-1 => x= 2
+)x+3=-3x-1 => x= \(\frac{-1}{2}\) ( không tmđk)
b)+c) Tương tự
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
Mình làm một vài câu thôi nhé, các câu còn lại tương tự.
Giải:
a) ??? Đề thiếu
b) \(\sqrt{-3x+4}=12\)
\(\Leftrightarrow-3x+4=144\)
\(\Leftrightarrow-3x=140\)
\(\Leftrightarrow x=\dfrac{-140}{3}\)
Vậy ...
c), d), g), h), i), p), q), v), a') Tương tự b)
w), x) Mình đã làm ở đây:
Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến
z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)
\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow4\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
- Câu a có chút thiếu sót, mong thông cảm :)
\(\sqrt{3x-1}\) = 4
Rút gọn :
\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|=\sqrt{5}-3+\sqrt{5}-2=2\sqrt{5}-5\)
Tính :
a) \(\sqrt{8,1.250}=\sqrt{81.25}=\sqrt{81}.\sqrt{25}=9.5=45\)
b) \(\sqrt{\dfrac{10.4,9}{16}}=\sqrt{\dfrac{1.49}{16}}=\dfrac{\sqrt{1}.\sqrt{49}}{\sqrt{16}}=\dfrac{7}{4}\)
c) \(\sqrt{8}.\sqrt{50}=\sqrt{8.50}=20\)
d) \(\dfrac{\sqrt{128}}{\sqrt{18}}=\sqrt{\dfrac{128}{18}}\approx2,7\)
\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =3-\sqrt{5}+\sqrt{5}-2=1\)
a)
Ta có :
\(\sqrt{\left(-8^2\right)}\) = \(\sqrt{64}\) = 8 vì 8 > 0 và 82 = 64
b)
Ta có :
\(\sqrt{16}\) = 4 vì 4 > 0 và 42 = 16
c)
Ta có :
\(\sqrt{1,44}\) = 1,2 vì 1,2 > 0 và ( 1,2 )2 = 1,44