giai ho mk nha
1200cc=?kg
giai thich ho mk nha
giai ho mk nha
giai ho mk nha
16x2+24xy+9y2=(4x)2+2*4x*3y+(3y)2=(4x+3y)2=[4x-(-3)y]2
=> giá trị của m là -3
\(16x^2+24xy+9y^2=\left(4x-my\right)^2\)
\(\Leftrightarrow\left(4x\right)^2+2\times4x\times3y+\left(3y\right)^2=\left(4x-my\right)^2\)
\(\Leftrightarrow\left(4x+3y\right)^2=\left(4x-my\right)^2\)
\(\Rightarrow m=-3\)
\(16x^2+24xy+9y^{^2}=\left(4x\right)^2+2.4x.3y+\left(3y\right)^2=\left(4x+3y\right)^2\)Dễ thấy -my=3y
=>-m=3
=>m=-3
giai ho mk nha
ta có:\(\dfrac{x-y}{z-y}\)=\(\dfrac{x-z+z-y}{z-y} \)=\(\dfrac{x-z}{z-y}\)+1=-10.
\(\Rightarrow\)\(\dfrac{x-z}{z-y} \)= -10 - 1= -11 \(\Rightarrow \) \(\dfrac{x-z}{z-y}\)= -11.
\(\Rightarrow\)-\((\dfrac{x-z}{z-y})\)=11 \(\Rightarrow\) \(\dfrac{x-z}{y-z}\)= 11.
Vậy kết quả là 11.
giai ho mk nha
Tứ giác MNPQ có MP vuông góc với NQ.
Ta có : MP = 12 cm, NQ = 8 cm.
\(\Rightarrow S_{MNPQ}=\frac{1}{2}\cdot MP\cdot NQ=\frac{1}{2}\cdot12\cdot8=48\left(cm^2\right)\)
Tứ giác MNPQ có MP vuông góc NP=> Đây là hình thoi
\(S_{MNPQ}=\frac{1}{2}.d_1.d_2=\frac{1}{2}.12.8=48cm^2\)
giai ho mk nha
Đặt A = \(\left(100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\right)-2\)
\(=\dfrac{\left(1+\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{101}{101}+\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{100\left(\dfrac{1}{101}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
= 100 - 2 = 98
giai ho mk nha
Ta có \(\frac{17}{3}=5+\frac{2}{3}\)\(=5+\frac{1}{1+\frac{1}{2}}\)
\(\Rightarrow m+\frac{1}{n+\frac{1}{p}}=\frac{17}{3}=5+\frac{1}{1+\frac{1}{2}}\)
\(\Leftrightarrow m=5,n=1,p=2\)
Vậy m=5, n=1, p=2
Giai ho mk nha
giai ho minh bai nay nha,mk tk cho:
a/x-1=5/-19
ngựa có 4 chân
tk mk nha
yêu mọi người nhiều lắm !!!!!!!!