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 !

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

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

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

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

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

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

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

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

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

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

đây là toán lớp 6 sao ?

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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

\(\)

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

a) \(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

\(\Rightarrow[-4x^2-\left(-4x\right).5]-[2x.8-\left(2x\right)^2]=-3\)

\(\Rightarrow[-4x^2-\left(-20x\right)]-\left(16x-4x^2\right)=-3\)

\(\Rightarrow\left(-4x^2\right)+20x-16x+4x^2=-3\)

\(\Rightarrow[\left(-4x^2\right)+4x^2]+\left(20x-16x\right)=-3\)

\(\Rightarrow4x=-3\)

\(\Rightarrow x=\frac{-3}{4}\)

b) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

\(\Rightarrow2^x+2^x.2+2^x.2^2+2^x.2^3=120\)

\(\Rightarrow2^x\left(1+2+4+8\right)=120\)

\(\Rightarrow2^x.15=120\)

\(\Rightarrow2^x=120:15=8=2^3\)

\(\Rightarrow x=3\)

Nhân hết ra,giải phương trình bậc cao đi