Mng ơi giúp minh câu 4 vs, mik đang cần gấp, cảm ơn mng!
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mng giúp mik bài 1 vs bài 3 vs, mik cần gấp lắm, cảm ơn mng.
0
TA
1
7 tháng 7 2023
2:
a: AC=căn 5^2-3^2=4cm
sin B=AC/BC=4/5
cos B=AB/BC=3/5
tan B=4/5:3/5=4/3
cot B=1:4/3=3/4
b: AB=căn 13^2-12^2=5cm
sin B=AC/BC=12/13
cos B=AB/BC=5/13
tan B=12/13:5/13=12/5
cot C=1:12/5=5/12
c: BC=căn 4^2+3^2=5cm
sin B=AC/BC=4/5
cos B=AB/BC=3/5
tan B=4/5:3/5=4/3
cot B=1:4/3=3/4
16 tháng 4 2023
IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d
XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is
C.
Bài 1
1. C
2. B
3. C
4. B
(Nên double-check trước khi chép)
J
3
Y
28 tháng 6 2023
\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 6 2023
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...
NM
1
7 tháng 11 2021
a) \(\Rightarrow\left|\dfrac{3}{4}+x\right|=0\Rightarrow\dfrac{3}{4}+x=0\Rightarrow x=-\dfrac{3}{4}\)
b) \(\Rightarrow x+0,4=\dfrac{4}{9}:\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow x=\dfrac{2}{3}-0,4=\dfrac{4}{15}\)
27 tháng 9 2021
\(a,\) Áp dụng t/c dtsbn:
\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}=\dfrac{5x}{50}=\dfrac{2z}{42}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=12\\z=42\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{124}{62}=2\\ \Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)
\(c,\) Áp dụng t/c dtsbn
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\\ \Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)
\(d,\) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(xy=54\Rightarrow2k\cdot3k=54\Rightarrow k^2=9\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=9\\x=-6;y=-9\end{matrix}\right.\)
\(e,\) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\Rightarrow x=5k;y=3k\)
\(x^2-y^2=4\Rightarrow25k^2-9k^2=4\Rightarrow16k^2=4\Rightarrow k^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2};y=\dfrac{3}{2}\\x=-\dfrac{5}{2};y=-\dfrac{3}{2}\end{matrix}\right.\)
\(f,\) Áp dụng t/c dtsbn:
\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)
\(\Rightarrow\left\{{}\begin{matrix}2x=y+z+1\\2y=x+z+1\\2z=x+y-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+z=3x-1\\x+y+z=3y-1\\x+y+z=3z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3x-1=\dfrac{1}{2}\\3y-1=\dfrac{1}{2}\\3z+2=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=-\dfrac{1}{2}\end{matrix}\right.\)
