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\(\frac{-5^3\cdot40\cdot4^3}{135\cdot\left(-2\right)^{14}\left(-100\right)^0}=\frac{-125\cdot2^3\cdot5\cdot\left(2^2\right)^3}{5\cdot27\cdot2^{14}\cdot1}=\frac{-125\cdot2^6}{27\cdot2^{11}}=\frac{-125}{27\cdot2^5}=\frac{-125}{864}\)
\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}\)\(=\frac{\left(-5\right)^3.5.2^3.2^6}{3^3.5.2^{14}.1}\)\(=\frac{-125}{864}\)
\(2a^3x^2y.8a^2x^3y^4.16a^3x^3y^3\)
\(=16^2.a^8.x^8.y^8\)
\(=\left(2axy\right)^8\)
\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
Nhận thấy \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
=> \(\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> \(2x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{6}\)
Vậy Min A = -1 <=> X = -1/6
a, \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\)
Dấu "=" xảy ra <=> 2x+1/3=0 <=> x= -1/6
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=\dfrac{86}{2}\\ x=43\)
\(105-\left[\left(2x+7\right)-13\right]=\left(-15\right)^{10}:\left(9^5.5^8\right)\\ 105-\left[\left(2x+7\right)-13\right]=15^{10}:3^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=5^{10}:5^8\\ 105-\left[\left(2x+7\right)-13\right]=25\\ \left(2x+7\right)-13=105-25\\ \left(2x+7\right)-13=80\\ 2x+7=80+13\\ 2x+7=93\\ 2x=93-7\\ 2x=86\\ x=86:2\\ x=43\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)
B1: a, |2 - x| + 2 = x
=> |2 - x| = x - 2
Dễ thấy (2 - x) và số đối của (x - 2)
=> |2 - x| = x - 2
=> 2 - x ≤ 0
=> x ≥ 2
b, Điều kiện: x + 7 ≥ 0 => x ≥ -7
Ta có: |x - 9| = x + 7
\(\Rightarrow\orbr{\begin{cases}x-9=x+7\\x-9=-x-7\end{cases}\Rightarrow}\orbr{\begin{cases}0x=16\left(loai\right)\\2x=2\end{cases}\Rightarrow x=1}\left(t/m\right)\)
\(\left(2x-5\right)^2=9\)
\(\left(2x-5\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-5=3\\2x-5=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-1\end{cases}}}\)
Học tốt ạ
\(\left(2x-5\right)^2=9\)
=> \(\orbr{\begin{cases}\left(2x-5\right)^2=3^2\\\left(2x-5\right)^2=\left(-3\right)^2\end{cases}}\)
=> \(\orbr{\begin{cases}2x-5=3\\2x-5=-3\end{cases}}\)
=> \(\orbr{\begin{cases}2x=8\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=1\end{cases}}\)