Tìm x, biết: (x+3)3 - x(3x+1)2 +x3 +1 = 54
tìm x biết:
(x+3)^3-x(3x+1)+(2x+1)(4x^2-2x+1)-3x^2=54
bn chép sai đề rùi, làm tui chết mệt, tui mới giải cho bn yến linh ( ở trên) bn xem nhé
e dap an cua cau nay la j vay:
x-2(x+3)=3x+54
Tìm x, biết:
a) (x-3)(x^2+ 3x +9)+x(2+x)(2x-x)=1
b) (x+3)^3 -x(3x+1)^2+(2x+1)(4x-2x+1)=54
tìm x, biết:
(x+3)^3 - 2x(3x+1)^2+(2x+1)(4x^2-2x+1)-3x^2=54
d) (2x – 1)2
e) (x + 6)3
f) (x + 3)(x2 – 3x + 9) – (54 + x3)
\(d,\left(2x-1\right)^2=4x^2-4x+1\\ e,\left(x+6\right)^3=x^3+18x^2+108x+216\\ f,\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)=x^3+27-54-x^3=-27\)
tìm x biết
a) (x-2)3-x(x+1)(x-1)+6x(x-3)=10
b) (x+1)3-(x-1)3-6(x-1)2= -10
c) x3+3x2+3x+28=0
d) x3-6x2+12x-7=0
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
Bài 1 :tìm x , biết :
(x-7)x+1 - (x-7)x+11 =0
Bài 2 :tìm x , biết :
a,|2x-3| > 5 c,|3x-1| ≤ 7 d,|3x-5| + |2x+3| = 7
Bài 3 :
a,tính tổng S = 1 + 52 + 54 + ....... + 5200.
b,so sánh 230 + 330 + 430 và 3.2410
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
Tìm x, biết :
a) (x-2)3 +6(x+1)2-x3+12=0
b) (x-5) (x+5) - (x+3)2+3(x-2)2=(x+1)2- (x+4)(x-4)+3x2
c) (2x+3)2 +(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)
d) (1-3x)2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
Giúp mk với ạ, mk cảm ơn !
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
c) (2x+3)2 +(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)
⇒4x2+12x+9+x2-1=5(x2+4x+4)-(x2+x-5x-5)+x+4
⇒5x2+12x+8=5x2+20x+20-x2-x+5x+5+x+4
⇒5x2+12x+8-5x2-20x-20+x2+x-5x-5-x-4=0
⇒x2-13x-21=0
Tìm x biết (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)-3x^2=54
Mình đang cần gấp các bạn giúp mình nhé!!
Bn thông cảm.Bài này mn ko bt làm
Viết theo HĐT
a) (x - 1)3
b) (2x + 3y)3
c) x3– 64.
d) 27x3+ 8y3
.
Bài 3. Tìm x, biết:
a) (x – 2)3– x2(x – 6) = 5
b) (x – 1)(x2+ x + 1) – x(x + 2)(x – 2) = 4
c) (x + 2)3– (x + 2) = 0
Bài 4. Tìm x biết (x + 2021)3+ (3x - 2022)3=(4x– 1)3
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
Tìm x biết :
(x+3)3 - x(3x + 1)2 +x3 +1 =54