(X+2/3) mũ 2= 25/9
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a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
=>-4x+2=0
hay x=1/2
c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\)
\(\left(x^2-9y^2\right)\left(4x+12y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-4\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x-3y-4\right)\)
\(3,\)
\(-x^2+2xy-y^2+25\)
\(=-\left(x^2-2xy+y^2\right)+25\)
\(=25-\left(x-y\right)^2\)
\(=5^2-\left(x-y\right)^2\)
\(=\left(5-x+y\right)\left(5+x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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23.27.24=23+7+4=214
23.24:25=23+4-5=27
(-3)+(-125)+(-25)
=-128+(-25)
=-153
25+(-38)=25-38=-13
126+159=285
120+(-135)+200
=120-135+200
=-15+200
=185
2^3x2^7=2^10x2^4=2^14=16384
2^3x2^4=2^7:2^5=2^2=4
1300-(150.2+(900+90):9)=1300-(150.2+990:9)=1300-(300+11)=1300-311=989
(-3) + (-125)+(-25)= -(3+125+25)= -153
25 + (-38)= -(38-25)= - 13
(126)+159=285
(120)+(-135)+200= -(135-120) + 200= (-15) + 200= +(200-15)=185
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)
\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)
\(=3^3-1+5^3-1\)
\(=27-1+125-1\)
\(=150\)
2 )
\(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b )
\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c )
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
=9x2^25-2^2x2^26/2^24x5^2-2^27x3
=9x2^25-2^28/2^24x5^2-2^27x3
=2^25x(9-2^3)/2^24x(5^2-2^3x3)
=2^25/2^24
=2^1=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
\(\left(x+\frac{2}{3}\right)^2=\frac{25}{9}=\left(\frac{5}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{5}{4}\\x+\frac{2}{3}=\frac{-5}{4}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{4}-\frac{2}{3}=\frac{7}{12}\\x=\frac{-5}{4}-\frac{2}{3}=\frac{-23}{12}\end{cases}}\)
\(\left(x+\frac{2}{3}\right)^2=\frac{25}{9}=\left(\frac{5}{4}\right)^2\)
\(\Leftrightarrow x+\frac{2}{3}=\frac{5}{4}\Leftrightarrow x=\frac{5}{4}-\frac{2}{3}=\frac{15-8}{12}=\frac{7}{12}\)
Vậy \(x=\frac{7}{12}\)