Mn làm từng bước giúp mk vs nhé
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Ta có:-(3-0,2.x)-80%=7,5
-3+0,2x-0,8=7,5
0,2x=7,5+3+0,8
X=11,3:0,2
X=56,5
Vậy x=56,5
Ta có:-(3-0,2.x)-80%=7,5
-3+0,2x-0,8=7,5
0,2x=7,5+3+0,8
X=11,3:0,2
X=56,5
Vậy x=56,5
3a + 9b = 138
= 3a + (3 x 3)b = 138
= 3a + 32 +b = 138
=3a + b + 2 = 138
còn lại cậu tự làm nhé
a) \(-x+80=-220-5x\)
\(\Rightarrow-x+5x=-220-80\)
\(\Rightarrow4x=-300\)
\(\Rightarrow x=-\dfrac{300}{4}\)
\(\Rightarrow x=-75\)
b) \(98+\left(x-12\right)+68=-80-x\)
\(\Rightarrow166+\left(x-12\right)=-80-x\)
\(\Rightarrow166+x-12=-80-x\)
\(\Rightarrow154+x=-80-x\)
\(\Rightarrow154+80=-x-x\)
\(\Rightarrow-2x=234\)
\(\Rightarrow x=-\dfrac{234}{2}\)
\(\Rightarrow x=-117\)
c) \(122+x-78=-55-4x-1\)
\(\Rightarrow44+x=-56-4x\)
\(\Rightarrow x+4x=-56-44\)
\(\Rightarrow5x=-100\)
\(\Rightarrow x=-\dfrac{100}{5}\)
\(\Rightarrow x=-20\)
d) \(663+9x=-x-37\)
\(\Rightarrow9x+x=-37-663\)
\(\Rightarrow10x=-700\)
\(\Rightarrow x=-\dfrac{700}{10}\)
\(\Rightarrow x=-70\)
a) \(\left(2x-4\right)\left(x-2\right)^3=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(x-2\right)^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\x-2-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
\(\Rightarrow x=2\)
b) \(3^{x-1}:81=3^3\)
\(\Rightarrow3^{x-1}:3^4=3^3\)
\(\Rightarrow3^{x-1-4}=3^3\)
\(\Rightarrow3^{x-5}=3^3\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=8\)
c) \(x^{13}=x\)
\(\Rightarrow x^{13}-x=0\)
\(\Rightarrow x\left(x^{12}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d) \(\left(x-9\right)^4=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^2=x-9\)
\(\Rightarrow\left(x-9\right)^2-\left(x-9\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-9-1\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)
\(=a-c+d-c+a-d\)
\(=2a-2c\)
b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)
\(=-a-b+c-d+a-b-c-d\)
\(=-2b-2d\)
c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)
\(=ab-ac-ad-ab-ac+ad\)
\(=-2ac\)
d) \(\left(a-b\right)+\left(c-d\right)+\left(a+c\right)-\left(b+d\right)\)
\(=a-b+c-d+a+c-b-d\)
\(=2a-2b+2c-2d\)
h) \(\left(a-b\right)-\left(c-d\right)-\left(a+d\right)-\left(b+c\right)\)
\(=a-b-c+d-a-d-b-c\)
\(=\left(a-a\right)-\left(b+b\right)-\left(c+c\right)+\left(d+d\right)\)
\(=-2b-2c\)
i) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)
\(=a-c+d-c+a-d\)
\(=\left(a+a\right)-\left(c+c\right)+\left(d+d\right)\)
\(=2a-2c\)
k) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)
\(=-a-b+c-d+a-b-c-d\)
\(=\left(-a+a\right)-\left(b+b\right)+\left(c-c\right)-\left(d+d\right)\)
\(=-2b-2d\)
n) \(-a\left(-b-c+d-e\right)-\left(-a\right)-\left(-b+c-d\right)\)
\(=ab+ac-ad+ae+a+b-c+d\)
Bài `4`
`1, 2y(x+2)-3x-6`
`=2y(x+2) -(3x+6)`
`=2y(x+2) -3(x+2)`
`=(x+2)(2y-3)`
`2, 3(x+4) -x^2-4x`
`=3(x+4)-(x^2+4x)`
`=3(x+4) -x(x+4)`
`=(x+3)(3-x)`
`3, 2(x+5) -x^2-5x`
`=2(x+5)-(x^2+5x)`
`=2(x+5)-x(x+5)`
`=(x+5)(2-x)`
`4, x^2 +6x-3(x+6)`
`= (x^2+6x) -3(x+6)`
`=x(x+6)-3(x+6)`
`=(x+6)(x-3)`
`5, x(x+y) -5x-5y`
`=x(x+y) -(5x+5y)`
`=x(x+y)-5(x+y)`
`=(x+y)(x-5)`
`6,x(x-y)+2x-2y`
`=x(x-y)+2(x-y)`
`=(x-y)(x+2`
a. Giả sử: \(3+\sqrt{12}>\sqrt{16}\)
<=> \(\sqrt{12}>1\) (thỏa mãn)
Vậy \(3+\sqrt{12}>\sqrt{16}\)
b. \(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
Ta thấy: 112 < 117
Vậy \(4\sqrt{7}< 3\sqrt{13}\)
Cảm ơn bn nhiều