tìm x biết:
-2x(x+3)+x(2x-1)=10
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1) |2x-1|=-19-x<=> \(\left[\begin{array}{nghiempt}2x-1=-19-x\\2x-1=19+x\end{array}\right.\)=> x=-6 hoặc x=20
2) |4-3x|=2x-10<=>\(\left[\begin{array}{nghiempt}4-3x=2x-10\\4-3x=10-2x\end{array}\right.\)=> x= 14/6 hoặc x=-6
3) |x|=3+2x<=> \(\left[\begin{array}{nghiempt}x=-3-2x\\x=3+2x\end{array}\right.\)=> x=-1 hoặc x=-3
1) - Nếu 2x - 1 < 0 thì -2x + 1 = -19 - x => -x = -20 => x = 20
- Nếu 2x - 1 > 0 thì 2x - 1 = -19 - x => 3x = -18 => x = -6
2) - Nếu 4 - 3x < 0 thì -4 + 3x = 2x - 10 => 6 = -x => x = -6
- Nếu 4 - 3x > 0 thì 4 - 3x = 2x - 10 => 14 = 5x => x = \(\frac{14}{5}\)
3) - Nếu x < 0 thì -x = 3 + 2x => -3x = 3 => x = -1
- Nếu x > 0 thì x = 3 + 2x => -x = 3 => x = -3
1.
\(\left|2x-1\right|=-19-x\)
\(2x-1=\pm\left(-19-x\right)\)
TH1:
\(2x-1=-19-x\)
\(2x+x=-19-1\)
\(3x=-20\)
\(x=-\frac{20}{3}\)
TH2:
\(2x-1=19+x\)
\(2x-x=19-1\)
\(x=18\)
Vậy x = -20/3 hoặc x = 18
2.
\(\left|4-3x\right|=2x-10\)
\(4-3x=\pm\left(2x-10\right)\)
TH1:
\(4-3x=2x-10\)
\(-3x-2x=-10-4\)
\(-5x=-14\)
\(x=\frac{14}{5}\)
TH2:
\(4-3x=-2x+10\)
\(-3x+2x=10-4\)
\(x=-6\)
Vậy x = 14/5 hoặc x = -6
3.
\(\left|x\right|=3+2x\)
\(x=\pm\left(3+2x\right)\)
TH1:
\(x=3+2x\)
\(x-2x=3\)
\(x=-3\)
TH2:
\(x=-3-2x\)
\(x+2x=-3\)
\(3x=-3\)
\(x=-1\)
1) \(\left|2x-1\right|=-19-x\)
\(=>\orbr{\begin{cases}-19-x=2x-1\\-19-x=-\left(2x-1\right)\end{cases}=>\orbr{\begin{cases}-19+1=2x+x\\-19-x=-2x+1\end{cases}}}\)
\(=>\orbr{\begin{cases}-18=3x\\-x+2x=1+19\end{cases}=>\orbr{\begin{cases}x=-6\\x=20\end{cases}}}\)
2) \(\left|4-3x\right|=2x-10\)
\(=>\orbr{\begin{cases}2x-10=4-3x\\2x-10=-\left(4-3x\right)\end{cases}=>\orbr{\begin{cases}2x+3x=4+10\\2x-10=-4+3x\end{cases}}}\)
\(=>\orbr{\begin{cases}5x=14\\-10+4=3x-2x\end{cases}=>\orbr{\begin{cases}x=\frac{14}{5}\\x=-6\end{cases}}}\)
3) \(\left|x\right|=3+2x\)
\(=>\orbr{\begin{cases}3+2x=x\\3+2x=-x\end{cases}=>\orbr{\begin{cases}2x-x=-3\\2x+x=-3\end{cases}}}\)
\(=>\orbr{\begin{cases}x=-3\\3x=-3\end{cases}=>\orbr{\begin{cases}x=-3\\x=-1\end{cases}}}\)
Ủng hộ mk nha ^_-
(x-2)2-(x-3)(x-3)=6
x2-2.x.2+22-x2-32=6
(x2-x2)-4x+(22+32)=6
-4x+13=6
-4x=6-13=-7
x=-7:(-4)=1,75
a/Ta có: M(x)+N(x) = (2x5 - 4x3 + 2x2 + 10x - 1) + (-2x5 + 2x4 + 4x3 + x2 + x - 10)
= 2x5 - 2x5 - 4x3 + 4x3 + 2x4 + 2x2 + x2 + 10x + x -1 - 10
= 2x4 + 3x2 + 11x - 11
b/ Ta có: A(x) = N(x)-M(x) = (-2x5 + 2x4 + 4x3 + x2 + x - 10) - (2x5 - 4x3 + 2x2 + 10x - 1)
= -2x5 - 2x5 + 2x4 + 4x3 + 4x3 + x2 - 2x2 + x - 10x -10 + 1
= -2x5 + 2x4 + 8x3 - x2 - 9x -9
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
-2x(x + 3) + x(2x - 1) = 10
-2x² - 6x + 2x² - x = 10
-7x = 10
x = -10/7