1/ Tìm x
a/ ( 2x - 1 ) 6 = ( 2x - 1 ) 8
b/ 5 x + 5 x + 2 = 650
c/ 3 x - 1 + 5 . 3 x - 1 = 162
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
b) 5(2x-3)+4x(x-2)+2x(3-2x)=0
\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
\(\Leftrightarrow x=\dfrac{15}{8}\)
vậy x=\(\dfrac{15}{8}\)
c)3x(2-x)+2x(x-1)=5x(x+3)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)
\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)
tìm x
a) (x+2)(x+3)-(x-2)(x+5)=6
b) (3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
5^x + 5^x+2=650
3^x-1 + 5×3^x-1=162
(2x-1)^6=(1-2x)^8
a)5x+5x+2=650
2*5x+2=650
2(5x+1)=650
5x+1=650/2
5x+1=325
5x=325-1
5x=324
=>x \(\in\phi\)
b)3x-1+5*3x-1=162
3x-1(1+5)=162
3x-1=162/6
3x-1=27=33
=>x-1=3
x=3+1
x=4
c)(2x-1)6=(1-2x)8
(2x-1)6=(-2x-1)8=(2x-1)8=(2x-1)6*(2x-1)2
=>(2x-1)2=1
2x-1=1
2x=1+1
2x=2
x=2/2
x=1
*)2x-1=0
2x=0+1
2x=1
x=1/2
tìm x
a,(x+1)^3-(x-1)^3-6(x-1)^2=-10
b,x(x+5)(x-5)-(x+2)(x^2-2x+4)=42
c,(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=49
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
a) (2x-1)3 = 27
(2x-1)3 = 93
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2
b) (2x-1)4 = 81
(2x-1)4 = (\(\pm\)34)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé
À phần b) bạn sửa dòng (2x-1)4 = (\(\pm\)34) thành (2x-1)4 = (\(\pm\)3)4 nhé
Mình vừa viết nhầm
TÌM X BIẾT:
a) 3 mũ -1 nhân 3 mũ x+5 nhân 3 mũ x-1 bằng 162
b) (2x-1)^6=(2x-1)^8
c) (2x-4)^8=512
Bài1:Rút gọn
a,(4x-5)(3x+2)-(7-3x)(x+2)
b,(-2x+1)(x-5)-3(x-2)(x+1)
c,(x^2-7)(x-5)+(3x^2+5)(2x-4)
d,(x^2+3x-2)(x+4)-4x(x-5)
Bài2:Tìm xbiết
a,(x-4)(x+3)-(x+1)(x-5)=8
b,(3x-2)(x+1)-3x(x+7)=13
c,(x+5)(x-5)-x(x+2)=9
d,(x-1)(x^2+x+1)-x(x^2-3)=1
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
7, Tìm x
a) \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\) b) |2x-5|=x+3
c) |2x+1|-2x=6 d) |x-5|+x=5
a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{8}\\x+\dfrac{1}{2}=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{8}\\x=\dfrac{-5}{8}\end{matrix}\right.\)
b) Ta có: \(\left|2x-5\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+3\left(x\ge\dfrac{5}{2}\right)\\2x-5=-x-3\left(x< \dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=3+5\\2x+x=-3+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)
1) tìm x
a) (5x+1)(x-4)-x+4=0
b)2x(x-5)-x(2x+3)=26
C) (x^2-x+1)(x+1)-x^3+3x=15
d) (x^2-5)(x+2)+5x=2x^2+17
Giải giúp mik với ak đang cần gấp
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)