Tính :
diện tích xung quanh hình hộp chữ nhật,hình lập phương
diện tích toàn phần hình hộp chữ nhật,hình lập phương
diện tích đáy hình hộp chữ nhật,hình lập phương
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)
mà \(\widehat{bMB}=50^0\)
nên \(\widehat{NMC}=50^0\)
Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)
=>\(\widehat{MNC}+110^0=180^0\)
=>\(\widehat{MNC}=70^0\)
Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)
=>\(\widehat{C}+50^0+70^0=180^0\)
=>\(\widehat{C}=60^0\)
b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)
=>\(\widehat{NMB}+50^0=180^0\)
=>\(\widehat{NMB}=130^0\)
Ta có: MN//AB
=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)
=>\(\widehat{CBA}=50^0\)
BN là phân giác của góc CBA
=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)
Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)
=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)
c: BN là phân giác của góc CBA
=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)
Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)
=>\(\widehat{BAN}+60^0+50^0=180^0\)
=>\(\widehat{BAN}=70^0\)
Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)
=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)
Bài 1:
a: \(\dfrac{a}{b}>1\)
=>\(\dfrac{a}{b}-1>0\)
=>\(\dfrac{a-b}{b}>0\)
mà b>0
nên a-b>0
=>a>b
b: a>b
=>\(\dfrac{a}{b}>\dfrac{b}{b}\)
=>\(\dfrac{a}{b}>1\)
c: a/b<1
=>\(\dfrac{a}{b}-1< 0\)
=>\(\dfrac{a-b}{b}< 0\)
mà b>0
nên a-b<0
=>a<b
d: a<b
=>\(\dfrac{a}{b}< \dfrac{b}{b}\)
=>\(\dfrac{a}{b}< 1\)
|x|+|y|<=3
mà x,y nguyên
nên \(\left(\left|x\right|;\left|y\right|\right)\in\left\{\left(0;3\right);\left(0;1\right);\left(0;2\right);\left(0;0\right);\left(1;1\right);\left(1;2\right);\left(3;0\right);\left(1;0\right);\left(2;0\right);\left(2;1\right)\right\}\)
=>(x;y)\(\in\){(0;0);(0;1);(1;0);(0;-1);(-1;0);(0;2);(2;0);(0;-2);(-2;0);(0;3);(0;-3);(3;0);(-3;0);(1;1);(1;-1);(-1;1);(1;2);(2;1);(-1;-2);(-2;-1);(1;-2);(-2;1);(-1;2);(2;-1)}
Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)
=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\widehat{BAC}\)
=>\(\widehat{IBC}+\widehat{ICB}=90^0-\dfrac{1}{2}\cdot\widehat{BAC}\)
Xét ΔBIC có \(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)
=>\(\widehat{BIC}+90^0-\dfrac{1}{2}\widehat{BAC}=180^0\)
=>\(\widehat{BIC}=180^0-90^0+\dfrac{1}{2}\cdot\widehat{BAC}=90^0+\dfrac{1}{2}\cdot\widehat{BAC}\)
\(-\dfrac{2}{5}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}\\ =\left(\dfrac{-2}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{3}{4}+\dfrac{1}{6}\right)\\ =\dfrac{-4}{5}+\left(\dfrac{9}{12}+\dfrac{2}{12}\right)\\ =\dfrac{-4}{5}+\dfrac{11}{12}\\ =\dfrac{-48}{60}+\dfrac{55}{60}\\ =\dfrac{55-48}{60}\\ =\dfrac{7}{60}\)
Bài 10:
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)
a:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)
\(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)
\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)
\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)
Bài 14:
x+y+z=0
=>x+y=-z; x+z=-y; y+z=-x
\(A=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\cdot\left(-x\right)\cdot\left(-y\right)\)
=-xyz
=-2
Bài 10:
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk\cdot k=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)
a:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=\left(\dfrac{dk\left(k^2+k+1\right)}{d\left(k^2+k+1\right)}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
b: Sửa đề: Chứng minh \(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{a}{d}\)
\(\dfrac{a^3+c^3+b^3}{c^3+b^3+d^3}=\dfrac{\left(dk^3\right)^3+\left(dk\right)^3+\left(dk^2\right)^3}{\left(dk\right)^3+\left(dk^2\right)^3+d^3}\)
\(=\dfrac{d^3k^9+d^3k^3+d^3k^6}{d^3k^3+d^3k^6+d^3}=\dfrac{d^3\cdot k^3\left(k^6+1+k^3\right)}{d^3\cdot\left(k^3+k^6+1\right)}=k^3\)
\(=\dfrac{dk^3}{d}=\dfrac{a}{d}\)
trả lời xong nhớ cho coin
ko bế ơi
like thôi