a)x^2+x+1/4
b)x^2-3x+9/4
c)x^2/4+x+1
d)x^2/4-1/2x+1/4
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1 tháng 9 2021 lúc 15:39
Bài 9: Tìm x, biết:
a)|-2x+1,5|=1/4
b)3/2-|1 1/4+3x|=1/4
c)|4x-1| - |3x-1/2|=0
d)|x-1|-2x=1/2
Giúp mình với mình đang cần gấp
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
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\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
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a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)
b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)
\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)
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3x^4 + 3x^2y^2 + 6x^3y - 27x^2
x^4 + x^3 - x^2 + x
2x^5 - 6x^4 - 2a^2x^3 - 6ax^3
x^5 + x^4 + x^3 + x^2 + x + 1
x^3 - 1 + 5x^2 - 5 + 3x - 3
1/4.(a + 1)^2 - 4/9.(a - 2)^2
12a^2b^2 - 3.(a^2b^2)^2
4x^2y^2 - (x^2 + y^2 - a^2)^2
(a + b + c)^2 + (a + b - c)^2 - 4c^2
x^3 - 1 + 5x^2 - 5 + 3x - 3
Phân tích đa thức sau thành nhân tử :
a)64x^4+y^4
b)x^8+x^4+1
c)x^3+2x^2+2x+1
d)x^4-2x^3+2x-1
Xem chi tiết
a)64x^4+y^4
b)x^8+x^4+1
c)x^3+2x^2+2x+1
d)x^4-2x^3+2x-1
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
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28 tháng 7 2021 lúc 9:29
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)Bài 1: a,√3x+4−√2x+1√x+3b, √2x−5+√x+2√2x+1c, √x+4−√1−x√1−2xd, √x+95−√2x+4Bài 2:a,√x+4√x+45x+2b, √x2−2x+1+√x2+4x+44c, √x+2√x−1+√x−2√x−12d,√x−2+√2x−5+√x+2+3√2x−57√2Bài 3:a, x2−7x6√x+5−30
Đọc tiếp
ai giúp mình giải bài này với được k mình đang cần gấp ( xin cảm ơn)
Bài 1:
a,√3x+4−√2x+1=√x+3
b, √2x−5+√x+2=√2x+1
c, √x+4−√1−x=√1−2x
d, √x+9=5−√2x+4
Bài 2:
a,√x+4√x+4=5x+2
b, √x2−2x+1+√x2+4x+4=4
c, √x+2√x−1+√x−2√x−1=2
d,√x−2+√2x−5+√x+2+3√2x−5=7√2
Bài 3:
a, x2−7x=6√x+5−30
Bài 3 Giải Phương Trình
a) 4x-2 = 1/x-1 - 5/x^2- x
b) -x^2+12x+4/x^2+3x-4 = 12/x+4 + 12/3x-3
c) 1/x-1 + 2/x^2-5 = 4/x^2+x+1
d) 1/2x^2+5-7 - 2/x-1 = 3/2x^2-5x-7
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
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12 tháng 4 2023 lúc 14:02
Giải pt:
a) | 5x | = 3x + 8
b) | -4x | = -2x + 11
c) | 3x - 1 | = 4x + 1
d) | 3 - 2x | = 3x - 7
e) 9 - | -5x | + 2x = 0
f) ( x + 1)² + | x + 10 | - x² - 12 = 0
g) | 4 - x | + x² - (5 + x)x = 0
h) | x - 1 | = | 2x - 3|
i) | x| + | x + 2 | = 4
k) | 2x + 1 | - | 5x - 2 | = 3
l) 2 | x | - | x + 3 | - 1 = 0
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
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Tìm x, biết: a) 121-(115+x)= 3x-(25-9-5x)-8
b)2x+2.3x+1.5x = 10800
c) (3|x-1/2) . (8/15-1/5)+2/3-1
d) x+1/2022 + x+2/2021= x+3/2020 + x+4/2019
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
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tìm x thỏa mãn:
a) (x+2)(x+3)-(x-2)(x-5)=-4
b) (x+1)(x2-x+1)-x(x-3)(x+3)=8
c) 4x2-9=(3x+1)(2x-3)
d) (3x+1)2-4(x-1)2=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x-5\right)=-4\)
\(\Leftrightarrow x^2+5x+6-x^2+7x-10=-4\)
\(\Leftrightarrow12x=0\)
hay x=0
b: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow x^3+1-x^3+9x=8\)
\(\Leftrightarrow9x=7\)
hay \(x=\dfrac{7}{9}\)
c: Ta có: \(4x^2-9=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x+1-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
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phân tích đa thức thành nhân tử
x^2 (x^4 - 1 ) ( x^2 + 2) + 1
9 ( x + 4/3 )( x + 2/3 )( x- 1/3 )( 1 -x ) - 4 ( x +1/3 )
2x^3 + 3x^2 + 8x - 5
( a + b - 3c )^2 ( a + b + 4c )^2 - 29c^2
2x3+3x2+8x-5=2x3-x2+4x2-2x+10x-5
=x2.(2x-1)+2x.(2x-1)+5.(2x-1)
= (2x-1).(x2+2x+5)
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Bài 2: Chứng to rằng các đa thức sau vô nghiệm:
a) f(x) = x +x+1
b) g(x) = x - x+1
c) mx)=(x-1)² +(x-2)
d) e(x) = |x-1+|x-2|
Bài 4: Tìm nghiệm của đa thức sau:
a) f(x)= x -2x-4
b) g(x) = x² + x +4
c) mx) = 8x - 12x +6x-2
d) n(x)= x+3x +3x+2
4:
a: f(x)=0
=>-x-4=0
=>x=-4
b: g(x)=0
=>x^2+x+4=0
Δ=1^2-4*1*4=1-16=-15<0
=>g(x) ko có nghiệm
c: m(x)=0
=>2x-2=0
=>x=1
d: n(x)=0
=>7x+2=0
=>x=-2/7
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