Tìm x, biết:
\(\left(2x+1\right)\times\left(3x-\frac{9}{2}\right)=0\)
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Tìm x biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
Rút gọn \(B=\left(x^4-x+\frac{x-3}{x^3+1}\times\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right)\times\frac{4x^2+6x+1}{\left(x+3\right)\left(4-x\right)}\)
Tìm x biết:
\(\left(2x+\frac{1}{3}\right)\left(\frac{3}{4}x-6\right)-\left(3x-\frac{2}{3}\right)\left(\frac{1}{2}x-\frac{6}{9}\right)=1\)
Tìm x, biếta,left(x^2+2xright)^2-2x^2-4x3b,left(x+frac{1}{2}right)^2-left(x+frac{1}{2}right)left(x+6right)8c,left(x+3right)^2-left(x+2right)left(x-2right)4x+17d,left(x-3right)left(x^2+3x+9right)-xleft(x^2-4right)1e,3x^2+7x10g,left(3x+5right)left(2x-1right)-6xleft(x+2right)xh,2left(x+3right)-x^2-3x0i,x^3-5x^2-14x0
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Tìm x, biết
a,\(\left(x^2+2x\right)^2-2x^2-4x=\)3
b,\(\left(x+\frac{1}{2}\right)^2-\left(x+\frac{1}{2}\right)\left(x+6\right)=8\)
c,\(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
d,\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-4\right)=1\)
e,\(3x^2+7x=10\)
g,\(\left(3x+5\right)\left(2x-1\right)-6x\left(x+2\right)=x\)
h,\(2\left(x+3\right)-x^2-3x=0\)
i,\(x^3-5x^2-14x=0\)
Bài 1 : tìm các giá trị của x biết :
a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
c) \(x^2=-6x-8\)
d) \(\frac{\left(x+1\right)^2}{3}-\frac{\left(x-2\right)^2}{3}=\frac{2x+1}{2}-\frac{\left(x-3\right)^2}{6}\)
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
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Tìm x , biết :
a. \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
b. \(2x^3-50x=0\)
c.\(5x^2-4\left(x^2-2x+1\right)-5=0\)
d. \(x^3-x=0\)
e. \(27x^3-27x^2+9x-1=1\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
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Tìm x:a) 3xleft(3x-8right)-9x^2+80b)6x-15-xleft(5-2xright)0c) x^3-16x0d) 2x^2+3x-50e) 3x^2-xleft(3x-6right)36f) left(x+2right)^2-left(x-5right)left(x+1right)17g) left(x-4right)^2-xleft(x+6right)9h) 4xleft(x-1000right)-x+10000i) x^2-360j) x^2y-2+x+x^2-2y+xy0k) xleft(x+1right)-left(x-1right).left(2x-3right)0l) 3x^3-27x0
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Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
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\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
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\(\left[2x+\frac{3}{5}\right]^2-\frac{9}{25}=0\)
\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\frac{9}{25}\)
\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\left[\frac{9}{25}\right]^2\)
\(\Rightarrow2x+\frac{3}{5}=\pm\frac{9}{25}\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{9}{25}\\2x+\frac{3}{5}=-\frac{9}{25}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{25}\\x=-\frac{12}{25}\end{cases}}\)
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Tìm x biết:
\(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
Nhận thấy x2 + 1 \(\ge\)1 > 0 \(\forall\)x
=> \(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
<=> \(\orbr{\begin{cases}2x^2-3=0\\3x^2-\frac{1}{0,12}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x^2=3\\3x^2=\frac{1}{0,12}\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=\frac{3}{2}\\x^2=\frac{1}{0,36}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{3}{2}}\\x=\pm\frac{1}{0,6}\end{cases}}\)
Vậy \(x\in\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}};-\frac{1}{0,6};\frac{1}{0,6}\right\}\)là giá trị cần tìm
\(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
Nhận thấy rằng x2 + 1 ≥ 1 > 0 ∀ x
=> \(\left(2x^2-3\right)\left(3x^2-\frac{1}{0,12}\right)\left(x^2+1\right)=0\)
<=> \(\orbr{\begin{cases}2x^2-3=0\\3x^2-\frac{1}{0,12}=0\end{cases}}\)
+) 2x2 - 3 = 0
<=> 2x2 = 3
<=> x2 = 3/2
<=> x = \(\pm\sqrt{\frac{3}{2}}\)
+) 3x2 - 1/0,12 = 0
<=> 3x2 - 25/3 = 0
<=> 3x2 = 25/3
<=> x2 = 25/9
<=> x = \(\pm\frac{5}{3}\)
Vậy S = { \(\pm\frac{5}{3}\); \(\pm\sqrt{\frac{3}{2}}\))