khi x>0, ta có

x - 38/7*x - 3/4 = 2*x + (-8/7)

-45/7*x=-11/28

x=-11/180( ko thoả )

khi x<0 có

-x -38/7x - 3/4 = -2x -8/7

bạn​ tự​ giải​ nhé​ rồi​ ktra dđiều​ kiện​ nhé !

\(E=\dfrac{4\left|x\right|+9}{\left|x\right|+1}\)

\(\left\{{}\begin{matrix} \left|x\right|\ge0\Rightarrow4\left|x\right|\ge0\Rightarrow4\left|x\right|+9\ge9\\\left|x\right|\ge0\Rightarrow x+1\ge1\end{matrix}\right.\)

\(MAX_E\Rightarrow MIN_{\left|x\right|+1}\)

\(MIN_{\left|x\right|+1}=1\)

\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)

\(\Rightarrow MAX_E=\dfrac{4.\left|0\right|+9}{\left|0\right|+1}=\dfrac{9}{1}=9\)

\(F=\dfrac{2\left|x\right|+8}{3\left|x\right|+1}\)

\(\left\{{}\begin{matrix}\left|x\right|\ge0\Rightarrow2\left|x\right|\ge0\Rightarrow2\left|x\right|+8\ge8\\\left|x\right|\ge0\Rightarrow3\left|x\right|\ge0\Rightarrow3\left|x\right|+1\ge1\end{matrix}\right.\)

\(MAX_F\Rightarrow MIN_{3\left|x\right|+1}\)

\(MIN_{3\left|x\right|+1}=1\)

\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)

\(\Rightarrow MAX_F=\dfrac{2.\left|0\right|+8}{3.\left|0\right|+1}=\dfrac{8}{1}=8\)

\(\)

A) (x—1)²= | 1/4–1/2–3/4 |

(x—1)²= | 1/4–2/4–3/4 |

(x—1)²=|—1|

(x—1)²=1

==> (x—1)=1 hoặc (x—1)=-1

        x=1+1 hoặc x—1=-1+1

       x=2 hoặc x=0

b)(x^x—8).(x²–15)<0

==> x^x—8 <0 và x²> 0

Hay x^x—8 >0 và x²<0

Mình chỉ biết tới đó thôi

\(a,\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

vậy__

b, k bt

đây là toán lớp 6 sao ?

\(\frac{7}{8}.(\frac{2}{12}+\frac{4}{10})\)

\(\Rightarrow\frac{7}{8}.(\frac{10+24}{60})\)

\(\Rightarrow\frac{7}{8}.\frac{34}{60}=\frac{238}{480}\)

bt2

\(2.x-\frac{5}{4}=\frac{20}{15}\)

\(\Leftrightarrow2x=\frac{20}{15}+\frac{5}{4}\)

\(\Leftrightarrow2x=\frac{80+75}{60}\)

\(\Leftrightarrow2x=2,5\)

\(\Leftrightarrow x=1,25\)

.7/8.(1/6+2/5)=7/8.17/30=119/240

3/2-5/6:1/4+\(\sqrt{4}\)=3/2-10/3+2=1/6

2x=20/15+5/4

2x=31/12

x=31/12:2

x=31/24

ko bt nha thông cảm

Mình bằng là:                                                                                                                                                                             x vẫn băng y nên:                                                                                                                                                                       x-(1+2+3+4)=1*2*3*4                                                                                                                                                                   x-10=24                                                                                                                                                                                      x=24+10                                                                                                                                                                                     x=34

a)\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)

\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)

\(\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)

\(\frac{1}{3}.x=\frac{35}{12}\)

\(x=\frac{35}{12}:\frac{1}{3}\)

\(x=\frac{35}{12}.\frac{3}{1}\)

\(x=\frac{35}{4}\)

b)\(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)

\(4,5.\left(0,1.x\right)=2,25.0,3\)

\(4,5.\left(0,1.x\right)=0,675\)

\(0,1.x=0,675:4,5\)

\(0,1.x=0,15\)

\(x=0,15:0,1\)

\(x=1.5\)

c)\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)

\(\left(\frac{1}{4}.x\right).2=8.0,02\)

\(\left(\frac{1}{4}.x\right).2=0,16\)

\(\frac{1}{4}.x=0,16:2\)

\(\frac{1}{4}.x=0,08\)

\(x=0,08:\frac{1}{4}\)

\(x=0,32\)

d)\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)

\(3.\left(6.x\right)=\frac{3}{4}.\frac{9}{4}\)

\(3.\left(6.x\right)=\frac{27}{16}\)

\(6.x=\frac{27}{16}:3\)

\(6.x=\frac{9}{16}\)

\(x=\frac{9}{16}:6\)

\(x=\frac{3}{32}\)

HỌC TỐT ^^

a.\(\left(3x-2\right)^2=16\)

Ta có: \(\left(3x-2\right)^2=16\)

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

\(\Rightarrow3x-2=4\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)

\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)

\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)

\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)

\(\Rightarrow x=\dfrac{7}{16}\)

c. \(5^{x+2}+5^x=3250\)

\(5^x.5^2+5^x=3250\)

\(5^x.\left(5^2+1\right)=3250\)

\(5^x.26=3250\)

\(5^x=125\)

\(5^x=5^3\)

\(x=5\)

\(1,\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-\left(2x^2-3x\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-2x^2+3x+3=0\)

\(\Rightarrow2x=-2\Rightarrow x=-1\)

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

\(\Rightarrow[\left(x^2\right)^2-\left(x-2\right)^2]-x^2\left(x^2-2\right)+8=0\)

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

\(\Rightarrow x^4-x^2+4x-4-x^4+2x^2+8=0\)

\(\Rightarrow x^2+4x+4=0\)

\(\Rightarrow\left(x+2\right)^2=0\Rightarrow x=-2\)