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a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
`|5x| = - 3x + 2`
Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :
`5x =-3x+2`
`<=> 5x +3x=2`
`<=> 8x=2`
`<=> x= 2/8=1/4` ( thỏa mãn )
Nếu `5x<0<=>x<0` thì phương trình trên trở thành :
`-5x = -3x+2`
`<=>-5x+3x=2`
`<=> 2x=2`
`<=>x=1` ( không thỏa mãn )
Vậy pt đã cho có nghiệm `x=1/4`
__
`6x-2<5x+3`
`<=> 6x-5x<3+2`
`<=>x<5`
Vậy bpt đã cho có tập nghiệm `x<5`
a: \(2x^3-50x=0\)
=>\(2x\left(x^2-25\right)=0\)
=>x(x-5)(x+5)=0
=>x∈{0;5;-5}
b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
=>2x(3x-5)+(3x-5)=0
=>(3x-5)(2x+1)=0
=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)
c: \(9\left(3x-2\right)=x\left(2-3x\right)\)
=>9(3x-2)-x(2-3x)=0
=>9(3x-2)+x(3x-2)=0
=>(3x-2)(x+9)=0
=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)
d: \(\left(2x-1\right)^2-25=0\)
=>(2x-1-5)(2x-1+5)=0
=>(2x-6)(2x+4)=0
=>(x-3)(x+2)=0
=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)
e: \(25x^2-2=0\)
=>\(25x^2=2\)
=>\(x^2=\frac{2}{25}\)
=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)
f: \(x^2-25=6x-9\)
=>\(x^2-6x-16=0\)
=>(x-8)(x+2)=0
=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)
g: 5x(x-3)-2x+6=0
=>5x(x-3)-2(x-3)=0
=>(x-3)(5x-2)=0
=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)
h: 3x(x-7)-2(x-7)=0
=>(x-7)(3x-2)=0
=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)
i: \(7x^2-28=0\)
=>\(7x^2=28\)
=>\(x^2=4\)
=>x=2 hoặc x=-2
j: 2x+1+x(2x+1)=0
=>(2x+1)(x+1)=0
=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)
k: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
=>(x+2)(x+2-x+2)=0
=>4(x+2)=0
=>x+2=0
=>x=-2
l: \(x^3+5x^2-4x-20=0\)
=>\(x^2\left(x+5\right)-4\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^2-4\right)=0\)
=>(x+5)(x-2)(x+2)=0
=>x∈{-5;2;-2}
m: \(x^2-25+2\left(x+5\right)=0\)
=>(x-5)(x+5)+2(x+5)=0
=>(x+5)(x-3)=0
=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)
n: \(x^2-3x+2=0\)
=>\(x^2-x-2x+2=0\)
=>x(x-1)-2(x-1)=0
=>(x-1)(x-2)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)
o: \(x^2-6x+8=0\)
=>\(\left(x-2\right)\left(x-4\right)=0\)
=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)
p: \(x^2-5x-14=0\)
=>\(x^2-7x+2x-14=0\)
=>(x-7)(x+2)=0
=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)
q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
=>\(x^2-4x+4-x^2+9=6\)
=>-4x+13=6
=>-4x=6-13=-7
=>x=7/4
r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)
=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)
=>-4x+26=18
=>-4x=-8
=>x=2
a) 3x - 2(5 + 2x) =45 - 2x
=> 3x - 10 - 4x = 45 - 2x
=> 3x - 4x + 2x = 45 + 10
=> x = 55
b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)
=> 5(2x + 17) = 3(x - 3)
=> 10x + 85 = 3x - 9
=> 7x = -94
=> x = -94/7
c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)
=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)
=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)
=> (-11x - 3).7 = (4x - 33).12
= -77x - 21 = 48x - 396
=> x = 3
d) (x - 1)(5x + 3) = (3x - 8)(x - 1)
=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0
=> (x - 1)(2x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\)
e) (x - 1)(x2 + 5x - 2) - (x3 - 1) = 0
=> (x - 1)(x2 + 5x - 2) - (x - 1)(x2 + x + 1) = 0
=> (x - 1)(4x - 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)
f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))
=> x = 50
b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)
\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)
f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)
\(\Leftrightarrow x=-66\)
a, (3x+1)(7x+3)=(5x-7)(3x+1)
<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0
<=> (3x+1)(7x+3-5x+7)=0
<=> (3x+1)(2x+10)=0
<=> 2(3x+1)(x+5)=0
=> 3x+1=0 hoặc x+5=0
=> x= -1/3 hoặc x=-5
Vậy...
a) (3x - 2)(4x + 5) = 0
⇔ 3x - 2 = 0 hoặc 4x + 5 = 0
1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3
2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4
Vậy phương trình có tập nghiệm S = {2/3;−5/4}
b) (2,3x - 6,9)(0,1x + 2) = 0
⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3
2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.
Vậy phương trình có tập hợp nghiệm S = {3;-20}
c) (4x + 2)(x2 + 1) = 0 ⇔ 4x + 2 = 0 hoặc x2 + 1 = 0
1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2
2) x2 + 1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)
Vậy phương trình có tập hợp nghiệm S = {−1/2}
d) (2x + 7)(x - 5)(5x + 1) = 0
⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2
2) x - 5 = 0 ⇔ x = 5
3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5
Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a) Ta có: \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Leftrightarrow\frac{x-17-33}{33}+\frac{x-21-29}{29}+\frac{x-2\cdot25}{25}=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)
nên x-50=0
hay x=50
Vậy: x=50
b) Ta có: \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)=2\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-4x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
hay \(x=\frac{41}{42}\)
a, \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=4-4\)
\(\Leftrightarrow\left(\frac{x-17-33}{33}\right)+\left(\frac{x-21-29}{29}\right)+\left(\frac{x-2.25}{25}\right)=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\) (*)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\Rightarrow\) Phương trình (*) xảy ra khi: \(x-50=0\Leftrightarrow x=50\)
Vậy phương trình có nghiệm duy nhất là x = 50.
Sao 17-33=50 vậy
☺