4x + 3x = 7014
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(x-25):4=17
x-25=17x4=68
x=68+25
x=93
x*5+x*2=7014
x*(5+2)=7014
x*7=7014
x=7014:7
x=1002
(x - 25) : 4 = 17
=> x - 25 = 68
=> x = 93
Vậy ....
x.5 + x . 2 = 7014
=> x( 5 + 2) = 7014
=> 7x = 7014
=> x = 1002
3075 - 9054 x ( 7014 : 7 - 1002 )
= 3075 - 9054 x ( 1002 - 1002 )
= 3075 - 9054 x 0
= 3075 - 0
= 3075
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
a: Ta có: \(\left(3x-1\right)^2-\left(3x+4\right)\left(3x-4\right)=32\)
\(\Leftrightarrow9x^2-6x+1-9x^2+16=32\)
\(\Leftrightarrow-6x=15\)
hay \(x=-\dfrac{5}{2}\)
b: Ta có: \(\left(4x+3\right)^2-\left(4x-1\right)\left(4x+1\right)=-14\)
\(\Leftrightarrow16x^2+24x+9-16x^2+1=-14\)
\(\Leftrightarrow24x=-24\)
hay x=-1
a: Sửa đề: 6x^3
\(\dfrac{6x^3-4x^2+3x-2}{3x-2}=\dfrac{2x^2\left(3x-2\right)+3x-2}{3x-2}=2x^2+1\)
b: \(\dfrac{6x^3-3x^2+4x+2}{3x^2+2}\)
\(=\dfrac{6x^3+4x-3x^2-2+4}{3x^2+2}\)
\(=2x-1+\dfrac{4}{3x^2+2}\)
a, \(\left(3x-x\right)^2\left(3x+1\right)\left(3x+1\right)=29\)
<=> \(4x^2\left(3x+1\right)^2=29\)
<=> \(4x^2;\left(3x+1\right)^2\inƯ\left(29\right)=\left\{\pm1;\pm29\right\}\)
4x^2 | 1 | -1 | 29 | -29 |
(3x+1)^2 | 29 | -29 | 1 | -1 |
x | 1/2 | ktm | \(\sqrt{\frac{29}{4}}\) | ktm |
x | \(\frac{\sqrt{29}-1}{3}\) | ktm | 0 | ktm |
b, Tương tự
b) ( 4x - 1 ) + ( 9 - 4x )( 3 + 4x ) = -8
<=> ( 4x - 1 ) + ( 27 + 24x - 16x2 ) = -8
<=> 4x - 1 + 27 + 24x - 16x2 = -8
<=> -16x2 + 28x + 26 = -8
<=> -16x2 + 28x + 26 + 8 = 0
<=> -16x2 + 28x + 34 = 0
<=> -2( 8x2 - 14x - 17 ) = 0
=> 8x2 - 14x - 17 = 0
\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=\left(\frac{-14}{2}\right)^2-\left(-17\right)\cdot8=185\)
\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt :
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)+\sqrt{185}}{8}=\frac{7+\sqrt{185}}{8}\)
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)-\sqrt{185}}{8}=\frac{7-\sqrt{185}}{8}\)
Lớp 7 mà nghiệm xấu nhỉ ?
4x + 3x = 7014
(4+3)x=7014
7x=7014
x=7014:7=1002
4x + 3x = 7014
( 4 + 3 ) x = 7014
7x = 7014
x = 7014 : 7
x = 1002