Câu 2:

a: x-158=32

=>x=158+32

=>x=190

b: \(x\cdot24=264\)

=>\(x=\dfrac{264}{24}\)

=>x=11

c: \(6x+9=3^7:3^4\)

=>\(6x+9=3^3\)

=>6x+9=27

=>6x=18

=>x=18/6=3

Câu 1:

a: \(86\cdot19+14\cdot19\)

\(=19\left(86+14\right)\)

\(=19\cdot100=1900\)

b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)

\(=4\cdot25-104\cdot25\)

\(=25\left(4-104\right)=-100\cdot25=-2500\)

c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)

\(=7\cdot2\cdot8\cdot5\)

\(=56\cdot10=560\)

d: \(59-\left[59+\left(-76\right)\right]\)

\(=59-59+76\)

=76

a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)

\(\Leftrightarrow9x+3=11\)

\(\Leftrightarrow9x=11-3\)

\(\Leftrightarrow9x=8\)

\(\Leftrightarrow x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)

\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)

\(\Leftrightarrow-62x+12=-50\)

\(\Leftrightarrow-62x=-50-12\)

\(\Leftrightarrow-62x=-62\)

\(\Leftrightarrow x=\dfrac{-62}{-62}\)

\(\Leftrightarrow x=1\)

a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(x^2+x+3x+3-x^2+5x=11\)

\(x+8x+3=11\)

\(x+8x=8\)

\(x\left(8+1\right)=8\)

\(x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)

\(-62x+12=-50\)

\(-62x=-62\)

\(x=1\)

\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)

Vậy \(x=12\).

(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12

Vậy x=12x=12.

cho tôi đúng đi

1.

6x + 1 ≥0

<=>6x≥-1

<=>x≥-1/6

2.

3x - 5 > 0 

<=> 3x > 5

<=> x > 5/3

3.

x - 7 > 0

<=> x > 7

4. 

-3x ≥0

<=>x≤0

5.

√5 - √3 . x ≥0

<=> √3 . x ≤ √5

<=> x ≤ √5/3 = (√15)/3

1: Ta có: \(4x^2-36=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

Nhiều thế ai làm được

mờ quá

sách j thế bạn?

a) 5x + 2x = 6^2 - 5^0

7x = 35

x=5

b) 6x + x = 5^11 : 5^9 + 3^1

7x = 28

x=4

c) 5x + 3x = 3^6 : 3^3 *4 +12

8x = 120

x=15

6x+x=5¹¹:5⁹+3¹

7x=25+3

7x=28

x=28/7

x=4

\(\Rightarrow7x=5^{11-9}+3\)

\(\Rightarrow7x=5^2+3\)

\(\Rightarrow7x=28\)

\(\Rightarrow x=\frac{28}{7}\)

6x+x=5¹¹:5⁹+3¹

6x+x=5²+3

6x+x=25+3

6x+x=28

x+x=28:6

2x=14/3

x=7/3