tính 3x+8 nếu x+1/x+4
tính 3x+8 nếu \(\frac{x+1}{x+4}=4\)
x+1/x+4=4
=> 4(x+4)=x+1
=>4x+16=x+1
=>4x-x+16-1=0
=>3x+15=0
=>3(x+5)=0
=>x+5=0
=>x=-5
Thay x=-5 vào 3x+8 ta được:
3x-5+8=-15+8
=-7
Chúc bạn học tốt
Từ \(\frac{x+1}{x+4}=4\) => x + 1 = 4 ( x + 4 )
<=> x + 1 = 4x + 16
<=> x - 4x = 16 - 1
<=> - 3x = 15
=> x = 15 : ( - 3 )
=> x = - 5
Thay x = - 5 vào 3x + 8 ta được :
3.( - 5 ) + 8 = - 15 + 8 = - 7
Vậy 3x + 8 = - 7 tại x = - 5
\(\frac{x+1}{x+4}\)=4
suy ra x+1=-4,x+4=-1
suy ra x=-5
suy ra 3x+8=-7
k cho mình nhé
Câu 1 : kết quả phép tính 3/4 + 1/4 . (-1)/2 là
A. -1/2
B. 5/8
C. -5/8
Câu 2 : nếu 3x = 5y và x - y = 8 thì
A. x = 20 ; y = -12
B. x = -20 ; y = 12
C. x = 20 ; y = 12
Câu 3 : nếu căn bậc x = 1/2 thì x mũ 2 bằng:
A. 1/2
B. 1/4
C. 1/8
Help pls
Câu 1: \(B.\frac{5}{8}\)
Câu 2: \(C.x=20;y=12\)
Câu 3: \(B.\frac{1}{4}\)
giá trị của biểu thức 3x+8 nếu \(\frac{x+1}{x+4}=4\)
\(\frac{x+1}{x+4}=4\)<=>\(\frac{x+1}{x+4}=\frac{4\left(x+4\right)}{x+4}\)
<=>\(x+1=4x+4\)<=>\(x=-1\) thay x=3 vào 3x+8=>\(3\times-1+8=5\)
Nếu \(\frac{x+1}{x+4}=4\)
<=> \(x=0\)
Thay x=0 vào BT ta có 3.0+8=0+8=8
tính :
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
2y-\(\frac{6xy+2y}{3x+2y}+\frac{2y-9x^2}{3x+2y}\)
Rút gọn rồi tính giá trị
c)C= 4(3x-2)2+(4-x)2-(6x-4)(8-2x) với x=149
d) (3x-4)2-9(x-2)(x+2) tại x=-2
e) x(x-3)2-(x-1)(x+5)-x(x-2)(x+2) tại x=-1
c) Ta có: \(C=4\left(3x-2\right)^2+\left(4-x\right)^2-\left(6x-4\right)\left(8-2x\right)\)
\(=4\left(9x^2-12x+4\right)+x^2-8x+16-\left(48x-12x^2-32+8x\right)\)
\(=36x^2-48x+16+x^2-8x+16-48x+12x^2+32-8x\)
\(=49x^2-112x+64\)
\(=\left(7x-8\right)^2\)
\(=\left(7\cdot149-8\right)^2\)
\(=1071225\)
d) \(\left(3x-4\right)^2-9\left(x-2\right)\left(x+2\right)\)
\(=9x^2-24x+16-9\left(x^2-4\right)\)
\(=9x^2-24x+16-9x^2+36\)
\(=-24x+52\)
\(=-24\cdot\left(-2\right)+52\)
=48+52=100
e) Ta có: \(x\left(x-3\right)^2-\left(x-1\right)\left(x+5\right)-x\left(x-2\right)\left(x+2\right)\)
\(=x\left(x^2-6x+9\right)-\left(x^2+4x-5\right)-x\left(x^2-4\right)\)
\(=x^3-6x^2+9x-x^2-4x+5-x^3+4x\)
\(=-7x^2+9x+5\)
\(=-7\cdot\left(-1\right)^2+9\cdot\left(-1\right)+5\)
\(=-7-9+5\)
=-16+5=-11
Tính giá trị biểu thức:
a) Q = (3x – 1)(9 x 2 – 3x + 1) – (1 – 3x)(1 + 3x + 9 x 2 ) tại x = 10;
b*) P = x 4 3 + y 2 3 biết xy = 4 và x + 2y = 8.
a) Rút gọn Q = 54 x 3 , thay x = 10 vào tính được Q = 54000;
b) Gợi ý x 4 + y 2 = x + 2 y 4 = 8 4 = 2 . Kết quả P = 2.
bài1: làm tính nhân ( rút gọn nếu có thể) a) (2x + 3)(x - 4) b) ( 3x + 1)(x - 2)
a: \(=2x^2-8x+3x-12=2x^2-5x-12\)
b: \(=3x^2-6x+x-2=3x^2-5x-2\)
Timf `x`:
`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)