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NL
1
15 tháng 12 2023
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
=>\(x^3+12x-8-x^3-125=11\)
=>12x-133=11
=>12x=144
=>\(x=\dfrac{144}{12}=12\)
PN
1
26 tháng 9 2021
a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)
\(\Leftrightarrow13x=13\)
hay x=1
TN
2
DH
26 tháng 7 2017
a, \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-x\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3-16x=5\)
\(\Rightarrow-16x-27=5\)
\(\Rightarrow-16x=32\Rightarrow x=-2\)
b, \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-5x^2+25x+5x^2-25x+125\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\Rightarrow12x=144\Rightarrow x=12\)
Chúc bạn học tốt!!!
26 tháng 7 2017
a)
\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3-3^3-x.\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3+16.x=5\)
\(\Rightarrow16x-27=5\)
\(\Rightarrow16x=32\)
\(\Rightarrow x=2\)
Vậy x = 2
b)
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\)
\(\Rightarrow12x=144\)
\(\Rightarrow x=12\)
Vậy x = 12
24 tháng 6 2016
+(x-3)2-x2=11
x2-6x+9-x2=11
-6x+9=11
-6x=2
x=2:-6
x=-1/3
(6x-3)2-36x(x-1)=40
36x2-36x+9-36x2+36x=40
9=40
=> đề sai
(2-x)2-7(x2+11)=0
4-4x+x2-7x2-77=0
-73-4x-6x2
ht bt
23 tháng 9 2021
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
23 tháng 9 2021
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
12 tháng 6 2018
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)
Vậy \(x=12\).
(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12
Vậy x=12x=12.
cho tôi đúng đi