60x8+60x6+60x9= ???
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\(\text{kết quả là :}\)
\(60\text{x 6}\)\(=360\)
\(\text{đáp số : 360}\)
\(thanks^{^{^{^{^{^{^{^{^{^6}}}}}}}}}\)
\(\frac{5}{12.17}+\frac{3}{34.10}+\frac{7}{60.9}+\frac{9}{27.36}\)
\(=\frac{5}{204}+\frac{3}{340}+\frac{7}{540}+\frac{9}{972}\)
\(=tự.tính.típ.nhé\)
a] [y+y] x 4 = 808
2yx4=808
2y=808:4
2y=202
y=202:2
y=101
b]y:6 = 60x6
y:6=360
y=360x6
y=2160
c]y:8 = 109[du 5]
y=109x8+5
y=877
d]677:y = 9 [du 2]
y=(677-2):9
y=75
tick mình nhé
a] [y+y] x 4 = 808
2yx4=808
2y=808:4
2y=202
y=202:2
x=101
b]y:6 = 60x6
y:6=360
y=360x6
x=2160
c]y:8 = 109[du 5]
y=109x8+5
y=872+5
y=877
d]677:y = 9 [du 2]
Cau 2
- TA CO :\(\frac{1}{A}\)+\(\frac{1}{B}\)+\(\frac{1}{c}\)=\(\frac{1}{2019}\)
- <=>\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=\(\frac{1}{a+b+c}\)
- <=>(\(\frac{1}{a}\)+\(\frac{1}{b}\))+(\(\frac{1}{c}\)-\(\frac{1}{a+b+c}\)=0
- <=>\(\frac{a+b}{ab}\)+\(\frac{a+b}{c\left(a+b+c\right)}\)=0
- <=>(a+b)(\(\frac{1}{ab}\)+\(\frac{1}{c\left(a+b+c\right)}\))=0
- <=>\(\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}\)=0
- <=>\(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
- gia su\(\left(a+b\right)=0\)=>c=2016 khi do
- \(\frac{a^{2019}+b^{2019}}{\left(ab\right)^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019+c^{2019}}}\)
- cac truong hop kia tuong tu
- \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
- <=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
- ,=>\(2\left(ab+bc+ac\right)=0\)
- <=>\(ab+bc+ac=0\)
- <=>\(ab=-\left(bc+ac\right),bc=-\left(ab+ac\right),ac=-\left(ab+bc\right)\)
- voi \(ab=-\left(bc+ac\right)\)=>\(c^2+2ab=c^2+ab-bc-ac=\left(a-c\right)\left(b-c\right)\)
- tuong tu \(a^2+2bc=\left(a-c\right)\left(a-b\right)\)
- \(b^2+2ac=-\left(b-c\right)\left(a-b\right)\)
- =>\(P=\frac{a^2}{\left(a-c\right)\left(a-b\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)\(=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
\(1:x< 0\left(B\right)\)
\(2:\left(D\right)\)
\(3:x< 2021\left(C\right)\)
\(4:x\ge15\left(D\right)\)
\(5:\)để pt có nghĩa thì 2x-5>0
\(2x>5< =>x>\frac{5}{2}\)
chọn (C)
\(6:\frac{1}{2}\sqrt{20}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(\frac{1}{2}\sqrt{20}-\sqrt{5}+2\)
\(\sqrt{5}-\sqrt{5}+2=2\)
chọn (B)
\(7:\frac{6xy^2}{x^2-y^2}\sqrt{\frac{\left(x-y\right)^2}{\left(3xy^2\right)^2}}\)
\(\frac{6xy^2}{x^2-y^2}\frac{x-y}{3xy^2}\)
\(\frac{2}{x+y}\)
chọn (B)
\(8:\left(1+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(\frac{3+\sqrt{3}}{\sqrt{3}+1}-1\right)\)
\(\left(1+\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-1\right)\)
\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(\sqrt{3}^2-1^2=3-1=2\)
chọn (D)
\(9:M=\left|1-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(M=\sqrt{3}-1+\sqrt{3}-1\)
\(M=2\sqrt{3}-2\)
chọn (A)
\(10:\sqrt{4+\sqrt{x^2-1}}=2\)
\(4+\sqrt{x^2-1}=2^2=4\)
\(\sqrt{x^2-1}=0\)
\(x^2-1=0< =>x=1\)
chọn (A)
a) Sửa đề: C/m tứ giác BEHC nội tiếp
Xét tứ giác BEHC có
\(\widehat{BEC}=\widehat{BHC}\left(=90^0\right)\)
\(\widehat{BEC}\) và \(\widehat{BHC}\) là hai góc cùng nhìn cạnh BC
Do đó: BEHC là tứ giác nội tiếp(Dấu hiệu nhận biết tứ giác nội tiếp)
Phép nhân hoá:
Ví dụ: Bác gấu đang bảo vệ những chú hươu khỏi đàn sói hung ác
60 x 8 + 60 x 6 + 60 x 9 = 60 x ( 8 + 6 + 9 )
= 60 x 23
= 1380
=60x(8+6+9)
=60x23
=1380
k đúng cho mk nhe