A\(^{A\left(x^{ }\right)=2x^3+5x^2-3x+1}\)
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17 tháng 9 2017
\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..
Câu sau cũng tương tự nhé
NL
26 tháng 2 2021
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(TH_1:3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(TH_2:-2x-7=0\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(TH_1:x=0\)
\(TH_2:x-1=0\)
\(\Leftrightarrow x=1\)
\(TH_3:2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)
\(TH_1:3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
\(TH_2:2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Rightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x-9x=-6-16+12\)
\(\Leftrightarrow11x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{11}\)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)
K
26 tháng 2 2021
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow3x+1=5x+8\)
\(\Leftrightarrow3x-5x=8-1\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(X=\dfrac{-7}{2}\)
b) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow9x^2-16-3x^2-4x=0\)
\(\Leftrightarrow6x^2-4x-16=0\)
\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)
\(\Leftrightarrow3x^2-6x+4x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Leftrightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x+16-12-9x+6=0\)
\(\Leftrightarrow11x+10=0\)
\(\Leftrightarrow x=\dfrac{-10}{11}\)
Vậy \(x=\dfrac{-10}{11}\)
HT
22 tháng 7 2017
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
HT
22 tháng 7 2017
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
TN
17 tháng 6 2019
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
28 tháng 5 2017
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
24 tháng 6 2018
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
19 tháng 8 2020
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
KN
19 tháng 8 2020
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
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