x=12

-4.|x-1| + (1/2-2,5)² = -3

-4.|x-1| + (-2)² = -3

-4.|x-1| + 4 = -3

-4.|x-1| = -3 - 4

-4.|x-1| = -7

|x-1| = (-7) : (-4)

|x-1| = 7/4

TH1: x - 1 = 7/4 => x = 7/4 + 1 = 11/4

TH2: 1 - x = 7/4 => x = 1 - 7/4 = -3/4

Vậy x = {11/4; -3/4}

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)

\(1-x=\frac{2}{5}:\frac{2}{5}=1\)

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)

\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)

học tốt nha

tích mình đi rùi mình giải

mk chi tim dc 1 nghiem la x=0

k biet con nghiem nao k

a) xy - 5y = 13

y . ( x - 5 ) = 13

Lập bảng ta có :

Vậy ( x ; y ) = ( 18 ; 1 ) = ( 6 ; 13 ) = ( -8 ; -1 ) = ( 4 ; -13 )

# Chúc bạn học tốt ^^!

a) xy - 5y = 13

y . ( x - 5 ) = 13

Lập bảng ta có :

Vậy ( x ; y ) = ( 18 ; 1 ) = ( 6 ; 13 ) = ( -8 ; -1 ) = ( 4 ; -13 )

thanks bn xênh gái nhìu!:>>>

Ta có: \(\frac{x+2}{3}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2\left(x+2\right)}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

\(\Rightarrow\frac{2x+4}{6}=\frac{y-1}{4}=\frac{z+5}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau được:

\(\frac{2x+4-\left(y-1\right)+z+5}{6-4+7}=\frac{2x+4-y+1+z+5}{6-4+7}=\frac{\left(2x-y+z\right)+\left(4+1+5\right)}{6-4+7}\)

                                                                                                     \(=\frac{17+10}{9}=\frac{27}{9}=3\)

Suy ra: \(2x+4=6.3\Rightarrow2x=14\Rightarrow x=7\)

            \(y-1=3.4\Rightarrow y=13\)

             \(z+5=3.7\Rightarrow z=16\)

Vậy x = 7 ; y = 13; z = 16

3/4+1/4:x=-3

1/4:x=(-3)-3/4

1/4:x=-15/4

x=-15/4.1/4

x=-15/16

đúng nha bn

3/4+3/4 : x= -3 

4/4 :x =-3

1:x =-3

x= -1/3

\(\frac{3}{4}+\frac{1}{4}:x=-3\)

=> \(\frac{1}{4}:x=-3-\frac{3}{4}=\frac{-15}{4}\)

=> \(x=\frac{1}{4}:\left(\frac{-15}{4}\right)=\frac{1}{4}.\left(\frac{-4}{15}\right)=\frac{-1}{15}\)

Vậy \(x=\frac{-1}{15}\)

\(2^x:1+2^x:2+...+2^x:49=2^{49}-1\)

\(2^x.1+2^x.\frac{1}{2}+...+2^x.\frac{1}{49}=2^{49}-1\)

\(2^x.\left(1+\frac{1}{2}+...+\frac{1}{49}\right)=2^{49}-1\)

Đặt: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)

=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\)

=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{49}}\right)\)

=> \(A=1-\frac{1}{2^{49}}=\frac{2^{49}-1}{2^{49}}\)

\(2^{x-1}+2^{x-2}+2^{x-3}+...+2^{x-49}=2^{49}-1\)

<=> \(\frac{2^x}{2}+\frac{2^x}{2^2}+\frac{2^x}{2^3}+...+\frac{2^x}{2^{49}}=2^{49}-1\)

<=> \(2^x\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\right)=2^{49}-1\)

<=> \(2^x.\frac{2^{49}-1}{2^{49}}=2^{49}-1\)

<=> \(2^x=2^{49}\)

<=> x = 49.

Lời giải:

a. $9x^2-16-(3x-4)(2x+5)=0$

$\Leftrightarrow [(3x)^2-4^2]-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4)-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4-2x-5)=0$

$\Leftrightarrow (3x-4)(x-1)=0$

$\Leftrightarrow 3x-4=0$ hoặc $x-1=0$

$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=1$.

b.

$x^2+4x=12$

$\Leftrightarrow x^2+4x-12=0$

$\Leftrightarrow (x^2-2x)+(6x-12)=0$

$\Leftrightarrow x(x-2)+6(x-2)=0$

$\Leftrightarrow (x-2)(x+6)=0$

$\Leftrightarrow x-2=0$ hoặc $x+6=0$

$\Leftrightarrow x=2$ hoặc $x=-6$

c.

$x^2-2x=35$

$\Leftrightarrow x^2-2x-35=0$

$\Leftrightarrow (x^2+5x)-(7x+35)=0$

$\Leftrightarrow x(x+5)-7(x+5)=0$

$\Leftrightarrow (x+5)(x-7)=0$

$\Leftrightarrow x+5=0$ hoặc $x-7=0$

$\Leftrightarrow x=-5$ hoặc $x=7$