Làm giúp mình bài 1 với ạ
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Bài 1:
Vì (d)//y=-2x+1 nên a=-2
Vậy: y=-2x+b
Thay x=1 và y=2 vào (d),ta được:
b-2=2
hay b=4
a, \(2x=5\Leftrightarrow x=\dfrac{5}{2}\)
b, \(2x-1=4x-8\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)
c, \(3x+9-6=2x+4\Leftrightarrow x=1\)
d, \(\left[{}\begin{matrix}2x+1=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
e, đk : x khác 0 ; 3
\(2x+8x-24=16\Leftrightarrow10x=40\Leftrightarrow x=4\left(tm\right)\)
Bài 2:
a: \(8x^3-36x^2+54x-27=\left(2x-3\right)^3\)
b: \(x^3+6x^2+12x+8=\left(x+2\right)^3\)
c: \(27x^3+8y^3=\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
d: \(x^3-\dfrac{y^3}{8}=\left(x-\dfrac{1}{2}y\right)\left(x^2+\dfrac{1}{2}xy+\dfrac{1}{4}y^2\right)\)
6) \(\dfrac{8^6}{256}=\dfrac{\left(2^3\right)^6}{2^8}=\dfrac{2^{18}}{2^8}=2^{10}=1024\)
7) \(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left[\left(\dfrac{1}{2}\right)^2\right]^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{40}=\left(\dfrac{1}{2}\right)^{55}=\dfrac{1}{2^{55}}\)
8) \(\left(\dfrac{1}{9}\right)^{25}\div\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{50}\div\left(\dfrac{1}{3}\right)^{30}=\left(\dfrac{1}{3}\right)^{20}=\dfrac{1}{3^{20}}\)
9)\(\left(\dfrac{1}{16}\right)^3\div\left(\dfrac{1}{8}\right)^2=\left(\dfrac{1}{2}\right)^{12}\div\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6=\dfrac{1}{64}\)
10) \(\dfrac{27^2.8^5}{6^2.32^3}=\dfrac{3^6.2^{15}}{3^2.2^2.2^{15}}=\dfrac{3^4}{2^2}=\dfrac{81}{4}\)
a) \(95.\left(-13\right)+\left(-13\right).15-13.\left(-10\right)\)
\(=13.\left(-95\right)+13.\left(-15\right)-13.\left(-10\right)\)
\(=13.\left[\left(-95\right)+\left(-15\right)-\left(-10\right)\right]\)
\(=13.\left(-90\right)\)
\(=-1170\)
3: \(\begin{cases}2x-3y=11\\ -4x+6y=5\end{cases}\Rightarrow\begin{cases}4x-6y=22\\ -4x+6y=5\end{cases}\)
=>\(\begin{cases}4x-6y-4x+6y=22+5\\ 2x-3y=11\end{cases}\Rightarrow\begin{cases}0x=27\\ 2x-3y=11\end{cases}\)
=>(x;y)∈∅
4: \(\begin{cases}3x+2y=1\\ 2x-y=3\end{cases}\Rightarrow\begin{cases}3x+2y=1\\ 4x-2y=6\end{cases}\)
=>\(\begin{cases}3x+2y+4x-2y=1+6=7\\ 2x-y=3\end{cases}\Rightarrow\begin{cases}7x=7\\ y=2x-3\end{cases}\)
=>\(\begin{cases}x=1\\ y=2\cdot1-3=2-3=-1\end{cases}\)
5: \(\begin{cases}2x+5y=2\\ 6x-15y=6\end{cases}\Rightarrow\begin{cases}6x+15y=6\\ 6x-15y=6\end{cases}\)
=>\(\begin{cases}6x+15y+6x-15y=6+6=12\\ 2x+5y=2\end{cases}=.\begin{cases}12x=12\\ 5y=2-2x\end{cases}\)
=>\(\begin{cases}x=1\\ 5y=2-2\cdot1=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=0\end{cases}\)
mình làm những bài bn chưa lm nhé
9B
10A
bài 2
have repainted
bàii 3
ride - walikking
swimming
watch










a) \(=5\sqrt{3}+9\sqrt{3}-12\sqrt{3}-8\sqrt{3}=-6\sqrt{3}\)
b) \(=8-3\sqrt{5}-\sqrt{45-2\sqrt{500}}=8-3\sqrt{5}-\left|5-2\sqrt{5}\right|=8-3\sqrt{5}-5+2\sqrt{5}=3-\sqrt{5}\)
c) \(\dfrac{8\sqrt{6}-12}{\sqrt{6}-4}-3\sqrt{\dfrac{2}{3}}+\dfrac{12}{\sqrt{6}-2}=\dfrac{\left(8\sqrt{6}-12\right)\left(\sqrt{6}+4\right)}{-10}-\sqrt{3}.\sqrt{2}+\dfrac{12\left(\sqrt{6}+2\right)}{2}=\dfrac{20\sqrt{6}}{-10}-\sqrt{6}+6\left(\sqrt{6}+2\right)=-2\sqrt{6}-\sqrt{6}+6\sqrt{6}+12=12+3\sqrt{6}\)