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TA CÓ: \(\left|x+1\right|=\frac{1}{2}\)
TH1: \(x+1=\frac{1}{2}\) TH2: \(x+1=\frac{-1}{2}\)
\(x=\frac{-1}{2}\) \(x=\frac{-3}{2}\)
THAY X =-1 /2 VÀO BIỂU THỨC M THAY X = -3/2 VÀO BIỂU THỨC M
\(M=5.\left(\frac{-1}{2}\right)^2-7.\frac{-1}{2}+\frac{1}{3}.\frac{-1}{2}-1\) \(M=5.\left(\frac{-3}{2}\right)^2-7.\frac{-3}{2}+\frac{1}{3}.\frac{-3}{2}-1\)
\(M=5.\frac{1}{4}-\frac{-1}{2}.\left(\frac{1}{3}+7\right)-1\) \(M=5.\frac{9}{4}-\frac{-3}{2}.\left(\frac{1}{3}+7\right)-1\)
\(M=\frac{5}{4}-\frac{-1}{2}.\frac{22}{3}-1\) \(M=\frac{45}{4}-\frac{-3}{2}.\frac{22}{3}-1\)
\(M=\frac{5}{4}-\frac{-11}{3}-1\) \(M=\frac{45}{4}-\left(-11\right)-1\)
\(M=\frac{47}{12}\) \(M=\frac{85}{4}\)
VẬY BIỂU THỨC M = 47/12 TẠI X= -1/2
................................= 85/4..............-3/2
CHÚC BN HỌC TỐT!!!!!!
=>(3x+2)(5x+1)=(5x+7)(3x-1)
(3x)(5x+1)+2(5x+1)=(5x)(3x-1)+7(3x-1)
15x2+3x+10x+2=15x2-5x+21x-7
(15x2-15x2)+(3x+10x+5x-21x)=-7-2
0-3x=-9
-3x=-9
x=(-9)/(-3)
x=3
\(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)
\(\Rightarrow\)\(\frac{3x+2}{3x-1}\)= \(\frac{5x+7}{5x+1}\)\(\Rightarrow\)1+\(\frac{3}{3x-1}\)=1+\(\frac{6}{5x+1}\)
\(\Rightarrow\)\(\frac{3}{3x-1}\)= \(\frac{6}{5x+1}\)\(\Rightarrow\)3.(5x+1) = 6.(3x-1)
\(\Rightarrow\)15x+3 = 18x -6
\(\Rightarrow\)3x = 6 +3
\(\Rightarrow\)x =3
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Rightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Rightarrow15x^2+10x+3x+2=15x^2-5x+21x-7\)
\(3x=9\)
\(x=3\)
Ta có: (3x+2)(5x+1)=(5x+7)(3x-1) (ĐKXĐ: x khác -7/5, x khác -1/5)
=> 15x2+13x+2=15x2+16x-7
=> 3x-9=0 =>x=3 (chọn)
ta có: f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)
\(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)
\(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)
\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)
Chúc bn học tốt !!!!!!
Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............
Ta có :\(\frac{\text{3x + 2}}{\text{5x + 7}}=\frac{\text{3x -1}}{\text{5x +1}}\)
=> ( 5x + 1 ) . ( 3x + 2 ) = ( 3x - 1 ) . ( 5x + 7 )
=> 5x(3x + 2 ) + ( 3x + 2 ) = 3x(5x + 7 ) - ( 5x + 7 )
=> ( 15x2 + 10x ) + ( 3x + 2 ) = ( 15x2 + 21x ) - ( 5x + 7 )
=> ( 3x + 2 ) + ( 5x + 7 ) = ( 15x2 + 21x ) - ( 15x2 + 10x )
=> ( 3x + 5x ) + ( 2 + 7 ) = ( 15x2 - 15x2 ) + ( 21x - 10x )
=> 8x + 9 = 11x
=> 9 = 11x - 8x
=> 9 = 3x
=> x = 3
Vậy x = 3
~~Học tốt~~
Ta có \(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)=\(\frac{3x+2-3x+1}{5x+7-5x-1}\)=\(\frac{1}{2}\)
=> \(\frac{3x+2}{5x+7}\)=\(\frac{1}{2}\)=> ( 3x + 2 ) . 2 = 5x + 7 . 1 => 6x + 4 = 5x + 7 => x=3
a ) \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=3x\left(5x+7\right)-\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
\(\Leftrightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow13x+2=16x-7\)
\(\Leftrightarrow13x-16x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Rightarrow x=3\)
b ) tương tự
3x+2 / 5x+7 = 3x-1 / 5x + 1
=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7)
=> 3x ( 5x+1) + 2( 5x+1) = 3x( 5x+7) - 5x - 7
15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7
15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2
13x - 16x = -9
-3x = -9
x = 3
Vậy x=3
Đúng 1000%
đề bài là j