Tìm x biết : a) | x - 1 | + | x - 5 | = 4x
b) | x + 2 | + | x + 4 | = 3x
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1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
A(x) = 7 - 3x + x2 + 4x - 1 - 3x2
= ( 7 - 1 ) + ( 4 - 3 )x + ( 1 - 3 )x2
= 6 + x - 2x2
B(x) = 2x - 4 - 2x2 - x + 5 - 3x
= ( -4 + 5 ) + ( 2 - 3 - 1)x - 2x2
= 1 - 2x - 2x2
Để A(x) = B(x)
=> 6 + x - 2x2 = 1 - 2x - 2x2
=> 6 - 1 = -x - 2x - 2x2 + 2x2
=> 5 = -3x
=> x = -5/3
Vậy x = -5/3
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\(A\left(x\right)=7-3x+x^2+4x-1-3x^2=6+x-2x^2\)
\(B\left(x\right)=2x-4-2x^2-x+5-3x=-2x+1-2x^2\)
Ta có : \(A\left(x\right)=B\left(x\right)\)
\(\Leftrightarrow6+x-2x^2=-2x+1-2x^2\)
\(\Leftrightarrow6+x-2x^2+2x-1+2x^2=0\)
\(\Leftrightarrow5+3x=0\Leftrightarrow3x=-5\Leftrightarrow x=-\frac{5}{3}\)
a)4×(x-5)-(x-1)×(4x-3)=5
=>4x-20-4x2+7x-3-5=0
=>-4x2+11x-28=0
=>-4(x2-\(\frac{11x}{4}\)+7)=0
=>\(-4\left(x-\frac{11}{8}\right)^2-\frac{327}{16}< 0\)
=>vô nghiệm
b) (3x-4)(x-2)=3x(x-9)-3
=>3x2-10x+8=3x2-27x-3
=>17x=-11
=>x=-11/17
c)(x-5)×(x-4)-(x+1)×(x-2)=7
=>x2-9x+20-x2+x+2=7
=>22-8x=7
=>-8x=-15
=>x=8/15
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
1. -4x( x + 3 )( x - 4 ) - 3x( x2 - x + 1 )
= -4x( x2 - x - 12 ) - 3x( x2 - x + 1 )
= -4x3 + 4x2 + 48x - 3x3 + 3x2 - 3x
= -7x3 + 7x2 + 45x
2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5
<=> 4x2 - 20x - ( 4x2 - 7x + 3 ) = 5
<=> 4x2 - 20x - 4x2 + 7x - 3 = 5
<=> -13x - 3 = 5
<=> -13x = 8
<=> x = -8/13
b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4
<=> 6( x2 - 7x + 12 ) - 6x2 + 12x = 4
<=> 6x2 - 42x + 72 - 6x2 + 12x = 4
<=> -30x + 72 = 4
<=> -30x = -68
<=> x = 34/15
Bài 1 :
\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)
\(=-7x^3+7x^2+45x\)
Bài 2 :
a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)
\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)
\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)
b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)
\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)
\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)
| x - 1 | + | x - 5 | = 4x (*)
+) Với x < 1
(*) <=> -( x - 1 ) - ( x - 5 ) = 4x
<=> 1 - x + 5 - x = 4x
<=> 6 - 2x = 4x
<=> 6 = 4x + 2x
<=> 6 = 6x
<=> x = 1 ( không thỏa mãn )
+) Với 1 ≤ x < 5
(*) <=> ( x - 1 ) - ( x - 5 ) = 4x
<=> x - 1 - x + 5 = 4x
<=> 4 = 4x
<=> x = 1 ( thỏa mãn )
+) Với x ≥ 5
(*) <=> ( x - 1 ) + ( x - 5 ) = 4x
<=> x - 1 + x - 5 = 4x
<=> 2x - 6 = 4x
<=> 2x - 4x = 6
<=> -2x = 6
<=> x = -3 ( không thỏa mãn )
Vậy x = 1
| x + 2 | + | x + 4 | = 3x (*)
+) Với x < -4
(*) <=> -( x + 2 ) - ( x + 4 ) = 3x
<=> -x - 2 - x - 4 = 3x
<=> -2x - 6 = 3x
<=> -2x - 3x = 6
<=> -5x = 6
<=> x = -6/5 ( không thỏa mãn )
+) Với -4 ≤ x < -2
(*) <=> -( x + 2 ) + ( x + 4 ) = 3x
<=> -x - 2 + x + 4 = 3x
<=> 2 = 3x
<=> x = 2/3 ( không thỏa mãn )
+) Với x ≥ 2
(*) <=> ( x + 2 ) + ( x + 4 ) = 3x
<=> x + 2 + x + 4 = 3x
<=> 2x + 6 = 3x
<=> 2x - 3x = -6
<=> -x = -6
<=> x = 6 ( thỏa mãn )
Vậy x = 6