\(a)\left(2x-1\right)^2+1=26\)

\(\left(2x-1\right)^2=25\)

\(TH1:2x-1=5\)

\(2x=6\)

\(x=3\)

\(TH2:2x-1=-5\)

\(2x=-4\)

\(x=-2\)

Vậy........

\(b)\left(2x-4\right)^3+2=66\)

\(\left(2x-4\right)^3=64=4^3\)

\(2x-4=4\)

\(2x=8\)

\(x=4\)

\(c)7^x+2+5.7^x+1+15=603\)

\(7^x\left(1+5\right)=603-15-1-2\)

\(7^x.6=585\)

Bạn xem lại phần này nhé . x tìm ra không được chẵn lắm á cậu.

Bạn xem lại câu a

a,(2x-1)²+1=26

=>(2x-1)²=26-1=25=5²

=>2x-1=5 và -5

=>x=(5+1):2=3 và x=(-5+1):2=-2

b,(2x-4)³+2=66

=>(2x-4)³=66-2=64=4³

=>2x-4=4

=>x=(4+4):2=4

c,7^x+2+5\(\cdot\)7^x+1+15=603

=>7^x\(\cdot\)7²+5\(\cdot\)7^x\(\cdot\)7=603-15=588

=>7^x(7²+5\(\cdot\)7)=588

=>7^x\(\cdot\)84=588

=>7^x=588:84=7

=>x=1

1,7y

Giúp mình vs.

\(\left|x-1\right|=2x-1\)

\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=-2x+1\end{cases}\left(đk:2x\ge\frac{1}{2}\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(tm\right)\end{cases}}\)

vậy \(x=\frac{2}{3}\)

mk làm lun nha

a, 2x^2-6x-3x-2x^2=26

-9x=26

x=-26/9

b,x^2+2.x.4+16-(x^2-1)=16

x^2+8x+16-x^2+1=16

8x=-1

x=-1/8

c,(2x)^2-2.2x.1+1-4(x^2-7^2)=0

4x^2-4x+1-4x^2+196=0

-4x=-197

x=197/4

d,x^2-5x-4x+20=0

-9x=-20

x=20/9 

**** cho mk nha

a) 2x (x - 5) - x (3 + 2x) = 26

=>  2x² - 10x - (3x - 2x²) = 26

=> 2x² - 10x - 3x - 2x² = 26

=> -13x = 26    => x = 26 : (-13) = -2

xin loi nhung hoi nhiu mik viet cau tra loi dc ko - Nguyễn Diệu Thảo

Bài 1.

a)

\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)

b)

\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

1a) (2x - 6)(x + 2) = 0

=> \(\orbr{\begin{cases}2x-6=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=6\\x=-2\end{cases}}\)

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

b) (x² + 7)(x² - 25) = 0

=> \(\orbr{\begin{cases}x^2+7=0\\x^2-25=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^2=-7\\x^2=25\end{cases}}\)

=>  x ko có giá trị vì x² \(\ge\)0 mà x²= -7

hoặc x = \(\pm\)5

suy ra 2x-6 =0 hoặc x+2=0

sau đó bạn giải từng trường hợp

1c) |2x - 1| = 4

=> \(\orbr{\begin{cases}2x-1=4\\2x-1=-4\end{cases}}\)

=> \(\orbr{\begin{cases}2x=5\\2x=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

vì x \(\in\)Z => ko có giá trị x

d) (x² - 9)(x² - 49) = 0

=> \(\orbr{\begin{cases}x^2-9=0\\x^2-49=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^2=9\\x^2=49\end{cases}}\)

=> \(\orbr{\begin{cases}x=\pm3\\x=\pm7\end{cases}}\)

nhìu dữ

a)3/2

b)-1/3

c)-5/6

d)0

e)-1/2

Bài 2

a=3

b=1/2

c=-1/3

d=0

e=9

f=-2/3

mk ko làm rõ đâu  nhe

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

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

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)