haha mấy bạn giải hay thật mjk mới đúng nè
(2x-3)2=9
(2x-3)2=32
=>3x-3=3 hoặc 3x-3=-3
3x=6 hoặc 3x=0
x=2 hoặc x=0
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Các câu hỏi tương tự
\(\left(x^2-x^3\right).\left(x-5\right)\)
\(\left(x+3\right)\left(x^2-3x+9\right)\)
\(\left(2x+y^2\right).\left(2x-y^2\right)\)
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
Rút gọn các biểu thức: Aleft(x+3right)left(x^2-3x+9right)-left(54+x^3right)Bleft(2x+yright)left(4x^2-2xy+y^2right)-left(2x-yright)left(4x^2+2xy+y^2right)Cleft(2x+1right)^2+left(1-3xright)^2+2left(2x+1right)left(3x-1right)Dleft(x-2right)left(x^2+2x+4right)-left(x+1right)^3+3left(x-1right)left(x+1right)
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Rút gọn các biểu thức:
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(\dfrac{9}{23}\left|2x-3\right|+\left(7y+17\right)^2=0\)
\(\left(x+2\right)^2=16\)
\(\left(2x-1\right)^3=8\)
\(\left(2x-3\right)^2=9\)
Tìm x:
1) \(\text{(x−1):0,16=−9:(1−x)}\)
2) \(\left(\left|x\right|-\dfrac{3}{2}\right)\left(2x^2-10\right)=0\)
3)\(8\sqrt{x}=x^2\left(x\ge0\right)\)
Tìm x,y biết
a)\(|x-3\left|+2\right|y+\frac{1}{2}|\le0\)
b)\(\left|3x+4\right|=2\left|2x-9\right|\)
c)\(\left|15x-1\right|\ge31\)
d)\(\left|9-7x\right|=5x-3\)
e)\(\left|x+3\right|-2x=\left|x-4\right|\)
\(a)\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)
\(b)\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)
\(_{c)\left(2x-3\right)^6=\left(2x-3\right)^6}\)
Bài 1: So sánh
a) \(-2^{30}\) và \(-3^{30}\)
b) \(35^5\) và \(6^{10}\)
Bài 2: Tính giá trị biểu thức
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
b) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(3^{2x-1}+2.9^{x-1}=405\)
\(\left(\dfrac{1}{3}\right)^{x-1}+5.\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)
\(\dfrac{3}{5}.\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}.\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)
Tìm x ( Giúp với mình cần gấp )