Xét ΔBAC có \(cosB=\frac{a^2+c^2-b^2}{2\cdot a\cdot c}\)

=>\(\left(4\sqrt2\right)^2+10^2-b^2=2\cdot4\sqrt2\cdot10\cdot cos45=8\sqrt2\cdot10\cdot\frac{\sqrt2}{2}=80\)

=>\(b^2=32+100-80=32+20=52\)

=>\(b=\sqrt{52}=2\sqrt{13}\)

Xét ΔABC có cos C=\(\frac{a^2+b^2-c^2}{2\cdot a\cdot b}\)

=>cosC=\(\frac{32+52-100}{2\cdot4\sqrt2\cdot2\sqrt{13}}=\frac{-16}{16\sqrt{26}}=-\frac{1}{\sqrt{26}}\)

=>\(\sin C=\sqrt{1-cos^2C}=\frac{5}{\sqrt{26}}\)

Diện tích tam giác CAB là:

\(S_{CAB}=\frac12\cdot CA\cdot CB\cdot\sin C\)

\(=\frac12\cdot\frac{5}{\sqrt{26}}\cdot2\sqrt{13}\cdot4\sqrt2=\frac{5\cdot2\cdot4}{2}=5\cdot4=20\)

Xét ΔABC có \(\frac{AB}{\sin C}=2R\)

=>\(2R=10:\frac{5}{\sqrt{26}}=\frac{10\sqrt{26}}{5}=2\sqrt{26}\)

=>\(R=\sqrt{26}\)

Ta có: \(S_{BCA}=\frac12\cdot AB\cdot AC\cdot\sin A\)

=>\(\frac12\cdot10\cdot2\sqrt{13}\cdot\sin A=20\)

=>\(\sin A=\frac{20}{10\sqrt{13}}=\frac{2}{\sqrt{13}}\)

\(S_{ACB}=\frac12\cdot BC\cdot h_{A}\)

=>\(\frac12\cdot4\sqrt2\cdot h_{A}=20\)

=>\(h_{A}=\frac{20}{2\sqrt2}=\frac{10}{\sqrt2}=5\sqrt2\)

a)

\(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{47}\\{35}\end{array}}\\\hline{\,\,\,\,82}\end{array}\)                                    \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{82}\\{47}\end{array}}\\\hline{\,\,\,35}\end{array}\)

\(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{526}\\{147}\end{array}}\\\hline{\,\,\,\,673}\end{array}\)                                \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{673}\\{147}\end{array}}\\\hline{\,\,\,526}\end{array}\)

b) 350 + 42 – 105 = 392 – 105 = 287

     1 000 – 300 + 77 = 700 + 77 = 777.

`# \text {Kaizu DN}`

`a)`

`(3x + 6) + (7x - 14) = 0?`

\(\Rightarrow3x+6+7x-14=0\\ \Rightarrow\left(3x+7x\right)+\left(6-14\right)=0\\ \Rightarrow10x-8=0\\ \Rightarrow10x=8\Rightarrow x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{4}{5}\)

Vậy, \(x=\dfrac{4}{5}\) 

`b)`

`17y + 35 + 4x + 17 = 42`

\(\Rightarrow\left(17y+17\right)+\left(35+4x\right)=42\\ \Rightarrow17\left(y+1\right)+\left(35+4x\right)=42\)

Bạn xem lại đề ;-;.

2:

a: 27(x-45)=0

=>x-45=0

=>x=45

b: (x-47)-115=0

=>x-47=115

=>x=162

d: x-105:21=15

=>x-5=15

=>x=20

1:

a: =35*(34+86)+65*(75+45)

=120*35+120*65

=120*100=12000

b: \(=39\left(53+47\right)-21\left(53+47\right)\)

=18*100=1800

ko bt

\(\dfrac{18}{45}=\dfrac{2}{5}\dfrac{63}{81}=\dfrac{7}{9}\) minh chi co the lam the nay thoi vi minh dang co viec gap

a, BCNN(210;280) = 840 => BC(210;280) = 840k (k ∈ N)

b, BCNN(105;135) = 945 => BC(105;135) = 945k (k ∈ N)

c, BCNN(24;36) = 72 => BC(24;36) = 72k (k ∈ N)

d, BCNN(48;80;72) = 720 => BC(48;80;72) = 720k (k ∈ N)

e, BCNN(42;55;91) = 30030 => BC(42;55;91) = 30030k (k ∈ N)

f, BCNN(12;36;102) = 612 => BC(12;36;102) = 612k (k ∈ N)

Tính nhanh:

     9 x 8 x 5 + 15 x 3 x 71 + 45

= 45 x 8 + 45 x 71 + 45

= 45 x ( 8 + 71 + 1 )

= 45 x 80

= 3600

 Tính tổng:

A = 115 + 110 + 105 + ... + 20 + 15

A= ( 15 + 115 ) x [ ( 115 - 15 ) : 5 + 1 ] 

A = 130 x 21

A = 2730

B = 99 - 97 + 95 - 93 + ... + 51 - 49

B = ( 99 - 97 ) + ( 95 -93 ) +...+ ( 51 - 49 )

B = 2 + 2 +...+ 2

B = 2 x 26

B = 52

C= 31 + 34 + 37 +...+ 91

C= ( 91 + 31 ) x [ ( 91 - 31 ) : 3 + 1 ]

C= 122 x 21

C= 2562

Ta có : 2 + 4 + 6 + ... + 2(n - 1) + 2n = 210

<=> 2[1 + 2 + 3 + ... + (n - 1) + n] = 210

<=> 1 + 2 + 3 + ... + n = 105

<=> [(n - 1) : 1 + 1)(n + 1) : 2 = 105

<=> n(n + 1) = 210

<=> n(n + 1) = 14.15

=> n = 14

Vậy n = 14

b) Ta có : 1 + 3 + 5 + ... + (2n - 1) = 225

<=> [(2n - 1 - 1) : 2 + 1](2n - 1 + 1) : 2 = 225

<=> n² = 225

<=> n² = 15²

<=> n = 15

Vậy n = 15

135/4

nha bạn