Bài làm:

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

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

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

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

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8

cmt gi may co bai tu di ma lam bat ng ta lam cho may chep ha

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

<=>\(2x=\frac{-2}{3}-\frac{-1}{2}=\frac{-1}{6}\)

<=>\(x=\frac{-1}{6}:2=-\frac{1}{2}\)

Vây \(x=-\frac{1}{2}\)

\(0.75-\left(-2x\right)=\frac{4}{5}\)

<=>\(\frac{3}{5}+2x=\frac{4}{5}\)

<=>\(2x=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}\)

<=>\(x=\frac{1}{5}:2=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

\(\left(2x+5\right)\left(1-x\right)=0\)

<=>\(2x+5=0\)hoặc \(1-x=0\)

<=>\(x=\frac{5}{2}\)hoặc \(x=1\)

Vậy \(x=\frac{5}{2}\)hoặc \(x=1\)

+) \(2x+\left(\frac{-1}{2}\right)=\frac{-2}{3}\)

\(\Rightarrow2x-\frac{1}{2}=\frac{-2}{3}\)

\(\Rightarrow2x=\frac{-2}{3}+\frac{1}{2}\)

\(\Rightarrow2x=\frac{-1}{6}\)

\(\Rightarrow x=\frac{-1}{6}:2\)

\(\Rightarrow x=\frac{-1}{12}\)

Vậy  \(x=\frac{-1}{12}\)

b) \(0,75-\left(-2x\right)=\frac{4}{5}\)

\(\Rightarrow\frac{3}{4}+2x=\frac{4}{5}\)

\(\Rightarrow2x=\frac{4}{5}-\frac{3}{4}\)

\(\Rightarrow2x=\frac{1}{20}\)

\(\Rightarrow x=\frac{1}{20}:2\)

\(\Rightarrow x=\frac{1}{40}\)

Vậy  \(x=\frac{1}{40}\)

c) \(\left(2x+5\right)\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+5=0\\1-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-5\\x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=1\end{cases}}\)

Vậy  \(x=\frac{-5}{2}\)hoặc \(x=1\)

_Chúc bạn học tốt_

a,(x-5)^2=25

(x-5)^2=5^2

=>x-5=5

x=5+5=10

Vậy x=10

b,(2x+1)^2=25

(2x+1)^2=5^2

=>2x+1=5

2x=6

x=6:2

x=3

Vậy x=3

c,(3x-2^4).7^3=2.7^4

3x-2^4=2.7^4:7^3=14

3x=16+14=30

x=30:3

x=10

Vậy x=10

d,2.3^x+3^2+x=891

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-7\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

b) \(x^2-2x=24\)

\(x^2-2x-24=0\)

\(\left(x-6\right)\left(x+4\right)=0\)

\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(5x^2+10x+10-5x^2+245=0\)

\(10x+255=0\)

\(x=-25.5\)

A) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)

th1\(\left(x-3\right)^2=2^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

th2: \(\left(x-3\right)^2=\left(-2\right)^2\)

\(\Rightarrow x-3=-2\)

\(\Rightarrow x=-2+3\)

\(\Rightarrow x=1\)

\(\Leftrightarrow x\in\left\{1;5\right\}\)

a) \(\left(x-3\right)^2-4=0\)

Ta xét 2TH:

TH1: \(x-3=2\)

\(x-3+3=2+3\)

\(x=5\)

TH2: \(x-3=-2\)

\(x-3+3=-2+3\)

\(x=1\)

=> \(x\in\left\{1;5\right\}\)

 Vì |x-1|+|x+1| luôn ko âm.

Với x âm .

=>2x-3 âm(loại)

Với x=1.

=?2x-3 âm (loại)

=>x>1.

=>|x-1|+|x+1|=x-1+x+1=2x=2x+3.

Hơi vô lí nhỉ!

Ta có: |x - 1| + |x + 1| = 2x - 3

Vì \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left|x+1\right|\ge0\end{cases}}\) \(\Rightarrow\) \(\left|x-1\right|+\left|x+1\right|\ge0\)\(\Rightarrow2x-3\ge0\)\(\Rightarrow2x\ge3\Rightarrow x\ge\frac{3}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1+x+1=2x-3\\x-1+x+1=-\left(2x-3\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=2x-3\\2x=-2x+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-2x=-3\\2x+2x=3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0x=-3\\4x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{3}{4}9\left(loai\right)\end{cases}}}\)

Vậy không có giá trị x thỏa mãn.

Đặt I x-1 I + I x+1I=A

lập bảng xét dấu:

- Xét x<-1

khi đó A= -(x-1)-(x+1)

             =-x+1-x-1

            =-(x+x)+(1-1)

             = -2x  +0=-2x

- Xét -1 <=  x <= 1

khi đó A= (x-1)-(x+1)

            =x- 1 - x + 1

            =(x-x)-(1-1)

             = 0-0=0

- Xét   x >= 1

khi đó A= (x-1)+(x+1)

            =x - 1 + x + 1

            =(x+x)-(1-1)

            =2x - 0 =2x

             => -2x= 2x - 3

                  -2x-2x=-3

                    -4x=3

                x= \(\frac{-4}{3}\)

             => 0= 2x - 3

                  2x-3=0

                 2x=3

                    x=\(\frac{3}{2}\)

             => 2x= 2x - 3

                  2x-2x=-3

                    0=3( Vô lí)

                x thuộc rỗng