2 nha bn

lớp 1 căng đét

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

c) (x+4)²-(x+1)(x-1)=16

⇒ x²+8x+16-x²+1=16

⇒ 8x+17=16

⇒ 8x=-1

⇒ x=-1/8

1.

$(x-2)(x-5)=(x-3)(x-4)$

$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)

Vậy pt vô nghiệm.

2.

$(x-7)(x+7)+x^2-2=2(x^2+5)$

$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$

$\Leftrightarrow -51=10$ (vô lý)

Vậy pt vô nghiệm.

3.

$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$

$\Leftrightarrow 4x+10=-8$

$\Leftrightarrow 4x=-18$

$\Leftrightarrow x=-4,5$

4.

$(x+1)^2=(x+3)(x-2)$

$\Leftrightarrow x^2+2x+1=x^2+x-6$

$\Leftrightarrow x=-7$ 

5.

$x^2-(2x-1)(x+3)=3-x(5+x)$
$\Leftrightarrow x^2-(2x^2+5x-3)=3-(5x+x^2)$
$\Leftrightarro -x^2-5x+3=3-5x-x^2$ (luôn đúng)

Vậy pt có nghiệm $x\in\mathbb{R}$

6.

$3(5-2x)-4(x+2)=5x-18$

$\Leftrightarrow 15-6x-4x-8=5x-18$

$\Leftrightarrow 7-10x=5x-18$

$\Leftrightarrow 25=15x$

$\Leftrightarrow x=\frac{5}{3}$

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2

2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2

7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7

3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0

8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9

4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2

9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9

5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8

10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4

1) 2(x + 5) + 3(x + 7) = 41

2x + 10 + 3x + 21 = 41

5x + 31 = 41

5x = 41 - 31

5x = 10

x = 10 : 5

x = 2

2) 5(x + 6) + 2(x - 3) = 38

5x + 30 + 2x - 6 = 38

7x + 24 = 38

7x = 38 - 24

7x = 14

x = 14 : 7

x = 2

3) 7(5 + x) + 2(x - 10) = 15

35 + 7x + 2x - 20 = 15

9x + 15 = 15

9x = 15 - 15

9x = 0

x = 0

4) 3(x + 4) + (8 - 2x) = 22

3x + 12 + 8 - 2x = 22

x + 20 = 22

x = 22 - 20

x = 2

5) 4(x + 5) + 3(7 - x) = 49

4x + 20 + 21 - 3x = 49

x + 41 = 49

x = 49 - 41

x = 8

6) 7(x - 1) + 5(3 - x) = 11x - 10

7x - 7 + 15 - 5x = 11x - 10

2x - 11x + 8 = -10

-9x = -10 - 8

-9x = -18

x = -18 : (-9)

x = 2

7) 4(2 + x) + 3(x - 2) = 12

8 + 4x + 3x - 6 = 12

7x + 2 = 12

7x = 12 - 2

7x = 10

x = 10/7

8) 5(2 + x) + 4(3 - x) = 10x - 15

10 + 5x + 12 - 4x = 10x - 15

10x - 15 = x + 22

10x - x = 22 + 15

9x = 37

x = 37/9

9) 7(x - 2) + 5(3 - x) = 11x - 6

7x - 14 + 15 - 5x = 11x - 6

11x - 6 = 2x + 1

11x - 2x = 1 + 6

9x = 7

x = 7/9

10) 5(3 - x) + 5(x + 4) = 6 + 4x

15 - 5x + 5x + 20 = 6 + 4x

6 + 4x = 35

4x = 35 - 6

4x = 29

x = 29/4

1: \(2^x=64\)

=>\(x=log_264=6\)

2: \(2^x\cdot3^x\cdot5^x=7\)

=>\(\left(2\cdot3\cdot5\right)^x=7\)

=>\(30^x=7\)

=>\(x=log_{30}7\)

3: \(4^x+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

=>\(2^x-1=0\)

=>\(2^x=1\)

=>x=0

4: \(9^x-4\cdot3^x+3=0\)

=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình sẽ trở thành:

\(a^2-4a+3=0\)

=>(a-1)(a-3)=0

=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)

=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)

=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)

=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)

=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)

=>\(3^{x+1}-2=0\)

=>\(3^{x+1}=2\)

=>\(x+1=log_32\)

=>\(x=-1+log_32\)

6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\) 

=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)

Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)

Phương trình sẽ trở thành:

\(\dfrac{1}{b}+b=2\)

=>\(b^2+1=2b\)

=>\(b^2-2b+1=0\)

=>(b-1)²=0

=>b-1=0

=>b=1

=>\(\left(2+\sqrt{3}\right)^x=1\)

=>x=0

7: ĐKXĐ: \(x^2+3x>0\)

=>x(x+3)>0

=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)

=>\(x^2+3x=4^1=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

a: x*3/4=1/5

=>x=1/5:3/4=1/5*4/3=4/15

b: x*3/7=2/5

=>x=2/5:3/7=2/5*7/3=14/15

c: 1/3+2/9=2/12x

=>1/6x=3/9+2/9=5/9

=>x=5/9*6=30/9=10/3

d: 4/15*x-2/3=1/5

=>4/15*x=2/3+1/5=10/15+3/15=13/15

=>4x=13

=>x=13/4

e: x:1/7=2/3

=>x=2/3*1/7=2/21

f: 1/9:x=7/3

=>x=1/9:7/3=1/9*3/7=3/63=1/21

j: 1/4+5/12=8/3:x

=>8/3:x=3/12+5/12=8/12=2/3

=>x=4

h: =>7/4:x=1/5+1/2=7/10

=>x=7/4:7/10=10/4=5/2

1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)

\(=x^2-2x+5-x^2-2x+7x-14\)

\(=3x-9\)

2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)

\(=-5x^2+25x+x^3-7x-3x^2+21\)

\(=x^3-8x^2+18x+21\)

3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^3-x^2-2x-x^2-4x+5\)

\(=x^3-2x^2-6x+5\)

Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!

ừ đúng dễ mà