Tìm x, biết:
a) (x- 25)-175 = 0;
b) 485 - ( 6.x + 60) = 5
c) 315 + (135 - x) = 450;
d) 346 + ( 210 - x) = 556.
Tìm x biết:
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
b) (2x - 1)2 - 3.(x - 2).(x + 2) - 25 = 0.
c) (x - 1)3 - x2.(x - 2) + 5 = 0.
d) x2 - 4x + 5 = 0.
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
b) (2x - 1)2 - 3.(x - 2).(x + 2) - 25 = 0.
<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0
<=> x2 - 4x - 12 = 0
<=> (x+2)(x-6) = 0
<=> x = -2 hoặc x = 6
d) x2 - 4x + 5 = 0.
<=> (x - 2)2 = -1 (vô lý)
Vậy phương trình vô nghiệm
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấppppppppp
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấpppppppppp
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Tìm x, biết:
a) 7x2 - 28 = 0
b) \(\dfrac{2}{3}\)x(x2 - 4) = 0
c) 2x(3x - 5) - (5 - 3x) = 0
d) (2x - 1)2 - 25 = 0
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
Tìm x biết:a) 29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29
b)x-10/30 + x-14/43 + x-5/95 + x-148/8 = 0
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
Tìm x, biết:
a) (x- 25)-175 = 0;
b) 485 - ( 6.x + 60) = 5
c) 315 + (135 - x) = 450;
d) 346 + ( 210 - x) = 556
a) x = 200.
b) x = 70.
c) x = 0.
d) x = 0.
tìm x biết:
a) x^2-2x+1=25
`x^2-2x+1=25`
`<=>(x-1)^2=25=5^2`
- TH1: `x-1=5<=>x=6`
- TH2: `x-1=-5<=>x=-4`
Vậy `S={6;-4}`
\(x^2-2x+1=25\)
\(\Leftrightarrow x^2-2.1.x+1^2=5^2\)
\(\Leftrightarrow\left(x-1\right)^2=5^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{6;-4\right\}\)
Tìm các số nguyên x, biết:
a, (22 + 5)(x2 + 25) = 0
b, (x2 + 7)(x2 - 49) < 0
c, (x2 - 7)(x2 - 49) < 0
d, (x2 - 36)(x2 - 81) ≤ 0
Tìm x biết: 420+65 . 4=(x + 175) : 5+ 30
882 : (2 . x - 15) = 18
x - 4867= ( 175 . 2050 . 70) : 25 + 23
134 - [ (96 - 7 . x) : 4-2]= 126
4 - {100-[ (100-5 . x) : 8-5] :10} : 25= 0
cần câu trả lời gấp, ai nhanh mik tick cho ạ
\(1) 420+65.4=(x+175):5+30\)
\(\implies 420+260=(x+175):5+30\)
\(\implies 680=(x+175):5+30\)
\(\implies 650=(x+175):5\)
\(\implies 650.5-175=x\)
\(\implies 3075=x\)
\(2)x-4867=(175.2050.70):25+23\)
\(\implies x-4867=1004500+23\)
\(\implies x-4867=1004523\)
\(\implies x=1009390\)
\(3) 134-[(96-7x):4-2]=126\)
\(\implies (96-7x):4-2=8\)
\(\implies(96-7x):4=10\)
\(\implies 96-7x=40\)
\(\implies 7x=56\)
\(\implies x=8\)
\(4) 4-\{100-[(100-5x):8-5]:10\}:25=0\)
\(\implies 4-\{100-[(100-5x):8-5]:10\}=0\)
\(\implies 100-[(100-5x):8-5]:10=4\)
\(\implies 100-[(100-5x):8-5]=40\)
\(\implies (100-5x):8-5=60\)
\(\implies (100-5x):8=65\)
\(\implies 100-5x=520\)
\(\implies 5x=-420\)
\(\implies x=-84\)
~ Hok tốt a~
ban nao biet thi minh h