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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
a)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
3x-1+5.3x-1=162
=>3x-1.(1+5)=162
=>3x-1.6=162=>3x-1=162:6=27=33
=>x-1=3=>x=4
5x+5x+2=650
=>5x+5x.52=650
=>5x.(1+25)=650
=>5x.26=650=>5x=650:26=25=52
=>x=2
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
a) \(3^x+5.3^x=45\)
\(3^x.1+5.3^x=45\)
\(3^x\left(1+5\right)=45\)
\(3^x.6=45\)
Xin lỗi mik nhấn nhầm
a) bn tính típ nhé
b) thiếu đầu bài
c) \(2^{3x+2}=4^{x+5}\)
bạn giải quyết nhé