\(2\left(x^2+8x+16\right)-x^2+4=0\)

\(\Leftrightarrow2x^2+16x+32-x^2+4=0\)

\(\Leftrightarrow x^2+16x+36=0\)

\(\Leftrightarrow x^2+16x+64=28\)

\(\Leftrightarrow\left(x+8\right)^2=28\)

\(\Leftrightarrow\orbr{\begin{cases}x_1=\sqrt{28}-8\\x_2=-\sqrt{28}-8\end{cases}}\)

\(2\left(x^2+8x+16\right)-x^2+4=0\)

\(2x^2+16x+32-x^2+4=0\)

\(x^2+16x+36=0\)

\(x^2+16x+64=28\)

\(\left(x+8\right)^2=28\)

bình phương thì chia lm 2 trường hợp 

lm tiếp phần sau 

a) ta có : (x-5)\(^2\) =x-5

\(\Rightarrow\)(x-5)\(^2\) - (x-5)=0

\(\Rightarrow\)(x-5)(x-6)=0

\(\Rightarrow\)\(\orbr{\begin{cases}x-5=0\\x-6=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

a)\(\left(x-5\right)^2=x-5\Leftrightarrow\left(x-5\right)^2-\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

b)\(\left|x-1\right|+x=1\)

TH1 Với \(x\ge1\Rightarrow x-1+x=1\Leftrightarrow x=1\left(tm\right)\)

TH2: Với \(x< 1\Rightarrow1-x+x=1\Leftrightarrow1=1\)luôn đúng. suy ra phương trình có nghiệm với mọi x<1.

Vậy phương trình đã cho có các nghiệm là \(x\le1.\)

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

2019x^2 - 2020x + 1 = 0

=> 2019x^2 - 2019x - x + 1 = 0

=> 2019x(x - 1) - (x - 1) = 0

=> (2019x - 1)(x - 1) = 0

=> 2019x - 1 =0  hoặc x - 1 = 0

=> 2019x = 1 hoặc x = 1

=> x = 1/2019 hoặc x = 1

a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

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

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

bấm chữ 0 đúng sẽ ra câu trả lời .

bấm vào chữ 0 đúng sẽ ra đáp án 

a)4x+4-3x+1=14

x+5=14

x=11

b)trường hợp 1  x²-9=0 

                        x²=9

->x=3;-3

-trường hợp 2: x+2=0

x=-2

c)-th1:x²+9=0

x²=-9

->x rỗng

d)xy+2x-y-2=0

(xy-y)+(2x-2)=0

y(x-1)+2(x-1)=0

(y+2)(x-1)=0

th1: y+2=0

y=-2

th2:x-1=0

x=1

(th1: trường hợp 1)

12 nguoi . 12 nguoi . Chuan do ban .

a, 11/12 - ( 2/5 + x ) = 2/3

<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)

b, 2x . ( x - 1/7 ) = 0

<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

vậy x={\(0;\frac{1}{7}\)}

c, 3/4 + 1/4 : x = 2/5

<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)

vậy x=-5/7

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)

\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)

\(\Leftrightarrow x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow4x=\frac{-20}{7}\)

\(\Leftrightarrow x=-\frac{5}{7}\)

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\)

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

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

\(x=-\frac{3}{20}\)

Vậy \(x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Rightarrow\begin{cases}x=0\\x=\frac{1}{7}\end{cases}\)

Vậy x = 0 hoặc \(x=\frac{1}{7}\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

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

\(\frac{1}{4}:x=-\frac{7}{20}\)

\(x=\frac{1}{4}:\left(-\frac{7}{20}\right)\)

\(x=-\frac{5}{7}\)

Vậy \(x=-\frac{5}{7}\)

Lời giải:

a.

PT $\Leftrightarrow 3x^2+\frac{x}{2}-3x^2+3x+2=0$
$\Leftrightarrow \frac{7}{2}x+2=0$
$\Leftrightarrow \frac{7}{2}x=-2$
$\Leftrightarrow x=-2: \frac{7}{2}=\frac{-4}{7}$
b.

PT $\Leftrightarrow 5x^2-3-5x^2-6x=0$

$\Leftrightarrow -3-6x=0$

$\Leftrightarrow 6x=-3$

$\Leftrightarrow x=\frac{-3}{6}=\frac{-1}{2}$