(7x-5)+2x=27
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21 tháng 2 2020
Mấy bài thực hiện phép tính bn tự làm nha
3.|x+4|-2.(x-1)=7-2x
3.|x-4|-2x-2=7-2x
=>2.|x-4| =7-2x+2x+2
=>2.|x-4| =9-4x
=>|x-4| =4,5-2x
=>x-4=4,5-2x hoặc x-4=-4,5+2x
2+2x=4,5+4 4,5-4=2x-2
3x =8,5 0,5 =x
x =8,5:3
x =2,833
Vậy x\(\in\){2,833;0,5}
Phần còn lại bn làm tương tự nha
Chúc bn học tốt
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
LV
18 tháng 1 2022
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
NN
3
25 tháng 9 2021
a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
c) \(\Rightarrow2^x=32\Rightarrow x=5\)
d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)
e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)
f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)
HP
25 tháng 9 2021
a. 2x . 4 = 128
<=> 2x + 2 = 27
<=> x + 2 = 7
<=> x = 5
b. (2x + 1)3 = 125
<=> (2x + 1)3 - 53 = 0
<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)
<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0
<=> (2x - 4)(4x2 + 14x + 31) = 0
<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)
c. 2x - 26 = 6
<=> 2x = 32
<=> x = 5
d. 64 . 4x = 45
<=> 43 . 4x = 45
<=> 43 + x = 45
<=> 3 + x = 5
<=> x = 2
e. 27 . 3x = 243
<=> 33 . 3x = 35
<=> 33 + x = 35
<=> 3 + x = 5
<=> x = 2
g. 49 . 7x = 2401 (Bn xem lại đề câu này)
<=> 72 . 7x = 74
<=> 72 + x = 74
<=> 2 + x = 4
<=> x = 2
TT
1
1 tháng 3 2020
\(2x^3-7x^2+7x-\frac{62}{27}=0\)
\(x^3+\left(x^3-7x^2+7x-\frac{62}{27}\right)=0\)
sau nx lần tra khảo idol
\(x^6-7x^5+7x^4-x^3.\frac{62}{27}=0\)
\(x=x^6-7x^5+7x^4-x^3.\frac{62}{27}\)
HẾT
NH
2
24 tháng 3 2020
\(1+\frac{2x}{x+4}+\frac{27}{2x^2+7x-4}=\frac{6}{2x-1}\left(x\ne-4;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow1+\frac{2x}{x+4}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6}{2x-1}=0\)
\(\Leftrightarrow\frac{2x^2+7x-4}{\left(x+4\right)\left(2x-1\right)}+\frac{2x\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{2x^2+7x-4}{\left(x+4\right)\left(2x-1\right)}+\frac{4x^2-2x}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6x+24}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{2x^2+7x-4+4x^2-2x+27-6x-24}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{6x^2-x-1}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow6x^2-x-1=0\)
\(\Leftrightarrow6x^2+2x-3x-1=0\)
<=> 2x(3x+1)-(3x+1)=0
<=> (3x+1)(2x-1)=0
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\left(tm\right)\\x=\frac{1}{2}\left(ktm\right)\end{cases}}}\)
Vậy pt có nghiệm \(x=\frac{-1}{3}\)
24 tháng 3 2020
\(ĐKXĐ:x\ne-4;x\ne\frac{1}{2}\)
\(1+\frac{2x}{x+4}+\frac{27}{2x^2+7x-4}=\frac{6}{2x-1}\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{2x\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}-\frac{6\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{2x^2+7x-4+4x^2-2x+27-6x-24}{\left(x+4\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow6x^2-x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow x=-\frac{1}{3}\)
12 tháng 1 2020
a) (2x-1)(x+2)=0
<=> 2x-1=0
Hoặc x+2=0
<=>2x=1
Hoặc x=-2
<=>x=1/2
Hoặcx=-2
Xl nha mk ko bt làm mấy câu kia
13 tháng 7 2021
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
13 tháng 7 2021
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
`(7x-5)+2x=27`
`7x-5+2x=27`
`(7x+2x)-5=27`
`9x-5=27`
`9x=27+5`
`9x=32`
`x=32:9`
`x=32/9`
Vậy `x=32/9`
(7x-5)+2x=27
\(\Rightarrow7x-5+2x=27\)
\(\Rightarrow9x-5=27\)
\(\Rightarrow9x=32\)
\(\Rightarrow x=\dfrac{32}{9}\)