2^7 : 2^2 + 5.2021^0 - 3.2^5 = ??
giải hộ em được không ạ ?
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a.(x+2)2-x(x+2)=0
\(\Leftrightarrow\)(x+2)(x-2-x)=0
\(\Leftrightarrow\)(x+2)*2=0
\(\Leftrightarrow\)x+2=0
\(\Leftrightarrow\)x=-2
vay s={-2}
b.\(\frac{2x+7}{3}\)-\(\frac{x-2}{4}\)=2
\(\Leftrightarrow\)\(\frac{4\left(2x+7\right)}{12}\)+\(\frac{-3\left(x-2\right)}{12}\)=\(\frac{24}{12}\)
\(\Leftrightarrow\)8x+28-3x+6=24
\(\Leftrightarrow\)5x=-10
\(\Leftrightarrow\)x=-2
vay s={-2}
c.|x+5|=3x+1
neu x+5\(\ge\)0 thi |x+5|=x+5
\(\Leftrightarrow\)x\(\ge\)-5
ta co phuong trinh
x+5=3x+1
\(\Leftrightarrow\)-2x=-4
\(\Leftrightarrow\)x=2( thoa man dieu kien x\(\ge\)-5)
neu x+5<0 thi |x+5|=5-x
\(\Leftrightarrow\)x<-5
ta co phuong trinh
5-x=3x+1
\(\Leftrightarrow\)-4x=-4
\(\Leftrightarrow\)x=1 (k thoa man dieu kien x<5)
vay s={2}
chuc bn hoc tot
\(a,\left(x+4\right)^2-x\left(x-5\right)=19\)
\(x^2+8x+16-x^2+5x=19\)
\(8x+5x=19-16\)
\(13x=3\)
\(x=\frac{3}{13}\)
\(b,x^2+3x-10=0\)
\(\Rightarrow x^2+5x-2x-10=0\)
\(\Rightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
\(a,\left(x+4\right)^2-x\left(x-5\right)=19\)
\(x^2+8x+16-x^2+5x=19\)
\(8x+5x=19-16\)
\(13x=3\)
\(x=\frac{3}{13}\)
\(b,x^2+3x-10=0\)
\(\Rightarrow x^2+5x-2x-10=0\)
\(\Rightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
a) (x+4)2 - x(x-5) = 19
x2+2.x.4+42 - x2 +5x = 19
8x +16 +5x =19
13x +16 =19
13x = 19-16=3
=> x=3:13=\(\frac{3}{13}\)
b) x2 +3x -10 =0
x(x+3) -10 =0
x(x+3) =10
=> x=2
chúc bn học tốt nha ^^ t chi mk nhé <3
\(\frac{x+2}{\left(x-2\right)^2}=\frac{16\left(x+2\right)}{16\left(x-2\right)^2}=\frac{16x+32}{16\left(x-2\right)^2}=\frac{x^2+12x+36-x^2+4x-4}{16\left(x-2\right)^2}\)
\(=\frac{\left(x+6\right)^2}{16\left(x-2\right)^2}-\frac{1}{16}\ge-\frac{1}{16}\)
Dấu = xảy ra khi x=-6
\(\dfrac{2^6.18+2^7}{2^6.5^2-2^6.3}=\dfrac{2^6\left(18+2\right)}{2^6.\left(5^2-3\right)}=\dfrac{20}{5^2-3}=\dfrac{20}{25-3}=\dfrac{20}{22}=\dfrac{10}{11}\)
\(=\dfrac{2^6.18+2^6.2}{2^6.25-2^6.3}=\dfrac{2^6\left(18+2\right)}{2^6\left(25-3\right)}=\dfrac{18+2}{25-3}=\dfrac{20}{22}=\dfrac{10}{11}\)
\(1+\dfrac{1}{2}.\dfrac{3.2}{2}+\dfrac{1}{3}.\dfrac{4.3}{2}+...+\dfrac{1}{500}.\dfrac{501.500}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{501}{2}\)
\(=\dfrac{2+3+4+...+501}{2}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}=62875\)
=-59 em nhé