Thực hiện phép tính
a) (√28 -2√14 +√7).√7+7√8
b) ( √8-3√2 +√10 ) (√2-3√0,4)
c) ( 15√50+ 5√200-3√450) : √10
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\(a.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=3.7-2.\sqrt{7.2.7}+14\sqrt{2}=21\) \(b.\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right).\dfrac{1}{10}=80\sqrt{2}.\dfrac{1}{10}=8\sqrt{2}\) \(c.\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{\dfrac{2}{5}}\right)=\left(\sqrt{5}-1\right)\left(2-6\sqrt{\dfrac{1}{5}}\right)\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
Mk ko ghi lại đề nhé
1 = \(350-4.19+4.7\)
\(=350-4.\left(19+7\right)\)
\(=350+4.26\)
\(=350-104=246\)
2 Câu này mình vẫn chưa hiểu là bạn ghi 27 3/5 tức là 27.3/5 hay 27\(\dfrac{3}{5}\)nên mk bỏ qua nhé
3 Cái đoạn 1 1/3 y chang câu 2
4 \(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}.\dfrac{12}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
5 \(=2^2+10^2-\left[9.\left(112:8\right)-11\right].1\)
\(=4+100-\left(9.14-11\right)\)
\(=104-115\)\(=-11\)
6 Đoạn 6 3/5 y chang 2,3
7 \(=\left(\dfrac{2}{3}-1\right)+\left(\dfrac{2}{7}-\dfrac{3}{7}\right)-\left(\dfrac{1}{14}-\dfrac{3}{28}\right)\)
\(=\dfrac{-1}{3}+\dfrac{-1}{7}-\dfrac{-1}{28}\)
\(=\dfrac{-196}{588}+\dfrac{-84}{588}-\dfrac{-21}{588}\)\(=\dfrac{259}{588}\)
Làm tạm tạm thôi xl vì có 3 câu ko hiểu
mik cách ra nghĩa là ghi 27, 3 phần 5 (mấy câu khác cx zị)hog phải nhân nha
@.@ Trời ơi, nhiều thế ^^
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)
\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)
b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
a: \(=\left(2\sqrt{7}+\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=3\sqrt{7}\cdot\sqrt{7}-2\sqrt{14}\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-2\sqrt{98}+14\sqrt{2}=21\)
b: \(=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-0,6\right)\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-0.6\right)\)
\(=-2+0.6\sqrt{2}+2\sqrt{5}-0.6\sqrt{10}\)
c: \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)