Tìm số nguyên x biết
m) (x-4).(3-x)=0
n) 2.(x-5)+4=2
h)2,5x-8=2x+3
bài 7 tìm x
1,x(x+3)-5(x+3)=0 2,5x(x-1)=x-1
3,(x+1)=(x+1)\(^2\) 4,x(2x-3)-2(3-2x)=0
5,\(\left(x-2\right)^2-4=0\) 6,\(36x^2=49\)
7,\(2x\left(x-6\right)-x+6=0\) 8,\(3x\left(2x-1\right)-24x+12=0\)
9,\(x^2-6x+8=0\) 10,\(x^2+2x-15=0\)
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
Bài 1: Cho 2 đa thức
M(x)=2,5x^2 -0,5x-x^3-1;1/2 N(x)=-x^3+2,5x^2-6+2x
a,Tìm A(x)=M(x) -N(x) .Rồi tìm nghiệm A(x)
b,Tìm đa thức B(x) biết B(x) =M(x)+N(x),tìm bậc của đa thức B(x)
Bài 3:Tìm nghiệm
a,f(x)=x^2-4x+3
b,f(x)=x^2-7x+12
c,f(x)=x^2+2x+1
d,f(x)=x^4+2
Ok help me pls ;-;
Bài 3:
a) Đặt f(x)=0
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
b) Đặt f(x)=0
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Bài 3:
c) Đặt f(x)=0
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
d) Đặt f(x)=0
\(\Leftrightarrow x^4+2=0\)
\(\Leftrightarrow x^4=-2\)(Vô lý)
Bài 1: Cho 2 đa thức
M(x)=2,5x^2 -0,5x-x^3-1;1/2 N(x)=-x^3+2,5x^2-6+2x
a,Tìm A(x)=M(x) -N(x) .Rồi tìm nghiệm A(x)
b,Tìm đa thức B(x) biết B(x) =M(x)+N(x),tìm bậc của đa thức B(x)
Bài 3:Tìm nghiệm
a,f(x)=x^2-4x+3
b,f(x)=x^2-7x+12
c,f(x)=x^2+2x+1
d,f(x)=x^4+2
Ok help me ;-;
bài 1: tìm số nguyên x, biết
a) \(\left(25-2x\right)^3\): 5 - \(3^2\)= \(4^2\) b) 2 x \(3^x\)= \(10\) x \(3^{12}\) + 8 x \(27^4\)
Lời giải:
a.
$(25-2x)^3:5-3^2=4^2$
$(25-2x)^3:5=4^2+3^2=25$
$(25-2x)^3=25.5=5^3$
$\Rightarrow 25-2x=5$
$\Rightarrow 2x=20$
$\Rightarrow x=10$
b.
$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$
$\Rightarrow 3^x=3^{14}$
$\Rightarrow x=14$
Bài 4: Tìm các số nguyên x biết
1) |x + 2| = 4
2) 3 – |2x + 1| = (-5)
3) 12 + |3 – x| = 9
4) |x + 9| = 12 + (-9) + 2
5) 2(4x – 8) – 7(3 + x) = |-4|(3 – 2)
6) 8(x – |-7|) – 6(x – 2) = |-8|.6 – 50
7) -7(5 – x) – 2(x – 10) = 15
8) 4(x – 1) – 3(x – 2) = -|-5|
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
1; |\(x\) + 2| = 4
\(\left[{}\begin{matrix}x+2=-4\\x+2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-4-2\\x=4-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- 6; 2}
1. Tìm x
a) 0,5x - 2 - ( 2,5x + 1) - (-x+2) = x
b) - 1/2 . (2x + 3) + 3/4 . (4x + 2/3) = 1/2 . ( 2- 3x)
c) 5^x + 5^x+2 = 650
tìm số nguyên x
5/x+1+4/x+1=3/-13
-x+2+2x+3+x+1/4+2x+1/6=8/3
3/2x+1+10/4x+2-6/6x+2=12/26
giúp mình mik đang vội]
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Leftrightarrow2x+1=13\)
hay x=6
1 tìm x biết :
a) -4/5 + 5/2x = -3/10
b)4/3 + 5/8 :x = 1/12 ( 4/3 hay 4/5 ý em k viết rõ nên bh k nhìn rõ ạ)
c) ( x - 1/3 ) ( x - 2/5 ) = 0
d) ( 3/4x - 9/16 ) ( 1,5 + -3/5 ) = 0
2: tìm x thuộc Z để :
a) 2x-1/x+2 nhận giá tị nguyên
b) x mũ 2 - 2x+1/ x+1 nhận gái trị nguyên
c) 7/ ( x+1) ( x+7) nhận giá trị nguyên
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5